Devu loves to play with binary strings a lot. One day he borrowed a binary string s of size n from his friend Churu. Before starting to play with it, he wants to make sure that string does not contain more than k consecutive equal characters. For achieving that, only kind of operation he is allowed to perform is to flip any i^th character of the string.

As Devu is always in hurry to meet his girlfriend, he wants you to help him in finding out the minimum number of operations he will need. Also he wants you to print one of the possible modified string too.

Input

• First line of input contains an integer T denoting the number of test cases.
• For each test case, there are two lines.
• First line contains two space separated integers n, k as defined in the problem.
• Next line contains string s of size n.

Output

• For each test case, print two lines.
• First line should contain an integer corresponding to minimum number of operations Devu needs.
• In second line, print one of the possible modified strings.

Constraints

Subtask #1: 20 points

• 1 ≤ T ≤ 100, 1 ≤ n ≤ 20, 1 ≤ k ≤ n

Subtask #2: 35 points

• 1 ≤ T ≤ 10², 1 ≤ n ≤ 10³, 1 ≤ k ≤ n

Subtask #3: 45 points

• 1 ≤ T ≤ 10⁵, 1 ≤ n ≤ 10⁵, 1 ≤ k ≤ n
• Sum of n over all the test cases is ≤ 10⁶

Example

Input:
3
2 1
11
2 2
11
4 1
1001

Output:
1
10
0
11
2
1010

Explanation

Example case 1: As 1 is occurring twice consecutively, we can convert 11 to 10 in a single operation.

Example case 2: You don‘t need to modify the string as it does not have more than 2 equal consecutive character.

Example case 3: As 0 is occurring twice consecutively, we can convert 1001 to 1010 in a two operations (Flip third and fourth character).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100000+1000;
int s[maxn];
int a[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,k;
        scanf("%d%d",&n,&k);
        char temp;
        getchar();
        for(int i=0; i<n; i++)
        {
            scanf("%c",&temp);
            s[i]=temp-'0';
        }
        int cur=1;
        int ans=0;
        int cnt=s[0];
        if(k!=1)
        {
            for(int i=1; i<n; i++)
            {
                if(s[i]==cnt)
                {
                    cur++;
                }
                else
                {
                    cur=1;
                    cnt=s[i];
                }
                if(cur==k+1&&i<n-1&&s[i+1]==cnt)
                {
                    s[i]=!s[i];
                    cur=0;
                    ans++;
                }
                if(cur==k+1&&i<n-1&&s[i+1]!=cnt)
                {
                    ans++;
                    s[i-1]=!s[i-1];
                    cur=1;
                }
                if(cur==k+1&&i==n-1)
                {
                    ans++;
                    s[i]=!s[i];
                }
            }
        }
        else
        {
            for(int i=1;i<n;i++)
            {
                if(s[i]!=(!s[i-1]))
                {
                    ans++;
                    s[i]=!s[i-1];
                }
            }
        }
        if(n-ans>=ans)
        {
        printf("%d\n",ans);
        for(int i=0; i<n; i++)
            printf("%d",s[i]);
        printf("\n");
        }
        else
        {
            printf("%d\n",n-ans);
            for(int i=0;i<n;i++)
            printf("%d",!s[i]);
            printf("\n");
        }
    }
    return 0;
}

The Chef once decided to prepare some nice dishes on his birthday. There are N items kept on his shelf linearly from position 1 to N. Taste of the i-th item is denoted by a integer A_i.

He wants to make Q dishes. A dish will be made using some ingredients in the continuous range A_L, A_{L + 1}, , , A_R (1-base indexing). Quality of the dish will be determined by the ingredient with minimum taste.

Chef wants help of his assistant Rupsa to find out sum and product of qualities of the dishes. As product of the qualities of the dishes could be very large, print it modulo 10⁹ + 7. Also, you are given an integerK and you are assured that for each dish, the size of continuous range of the ingredients (i.e. R - L + 1) will always lie between K and 2 * K, both inclusive.

Method of generation of Array A 
You are given non-negative integer parameters a, b, c, d, e, f, r, s, t, m, A[1]

for x = 2 to N:
	if(t^x mod s  <= r)        // Here t^x signifies "t to the power of x"
		A[x] = (a*A[x-1]^2 + b*A[x-1] + c) mod m
	else
		A[x] = (d*A[x-1]^2 + e*A[x-1] + f) mod m

Method of generation of range of ingredients for Q dishes 
You are given non-negative integer parameters L1, La, Lc, Lm, D1, Da, Dc, Dm

for i = 1 to Q:
	L1 = (La * L1 + Lc) mod Lm;
	D1 = (Da * D1 + Dc) mod Dm; 
	L = L1 + 1;
	R = min(L + K - 1 + D1, N);

Input

• The first line contains three integers N, K and Q.
• The second line contains the integers a, b, c, d, e, f, r, s, t, m, and A[1].
• Then third line contains the integers L1, La, Lc, Lm, D1, Da, Dc, and Dm

Output

Output two space separated integers:

• The sum of qualities of the dishes.
• The product of qualities of the dishes modulo 10⁹+7.

Constraints

• 1 ≤ N, Q ≤ 10⁷
• 1 ≤ K ≤ N
• 0 ≤ a, b, c, d, e, f, r, s, t, m, A[1] ≤ 10⁹+7
• 1 ≤ Lm ≤ N - K + 1
• 1 ≤ Dm ≤ K + 1
• 1 ≤ La, Lc ≤ Lm
• 1 ≤ Da, Dc ≤ Dm
• 1 ≤ L1 ≤ N
• 1 ≤ D1 ≤ K

Sub tasks

• Subtask #1: 1 ≤ N, Q ≤ 1000 (10 points)
• Subtask #2: 1 ≤ Q ≤ 10⁴ (20 points)
• Subtask #3: original constraints (70 points)

Example

Input:
4 2 1
1 1 1 1 1 1 1 1 1 100 1 
1 1 1 3 1 1 1 2

Output:
13 13

Explanation

• The array A comes out to be {1, 3, 13, 83} and the first dish has L = 3 and R = 4. The minimum in this range is 13, thus the sum and product both are 13 and hence the answer.

Note

Multiplier for C# and Java have been reduced to 1.5 instead of 2.