2
3
2 1 1
2 2 3
5
5 4 3 2 1
1 2 3 4 5

1
0

Hint

In the first example, there is only one valid permutation (2,1,3) .

In the second example, it is obvious that there is no valid permutation.
 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm> 
using namespace std;
typedef long long LL;
const int mod =998244353;
const int N=100005;
int b[N];
int c[N];
LL dp[N];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
    	int n;
    	scanf("%d",&n);
    	int Min=999999;
    	int Max=-1;
    	int flag=1;
    	for(int i=1;i<=n;i++)
    	{
    		scanf("%d",&b[i]);
    		if(b[i]>Min) flag=0;
    		if(b[i]<1||b[i]>n) flag=0;
    		Min=min(Min,b[i]);
		}
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&c[i]);
			if(c[i]<Max)flag=0;
			if(c[i]<1||c[i]>n) flag=0;
			if(c[i]<b[i]) flag=0;
			Max=max(Max,c[i]);
			
		}
		if(!flag||c[1]!=b[1]) puts("0"); //前 1个的最大最小值一定相等 ，都是这个数嘛 
		else
		{
			memset(dp,0,sizeof(dp));
			dp[1]=1;
			int num=1;
			for(int i=2;i<=n;i++) //dp核心 
			{
				if(c[i]==c[i-1]&&b[i]==b[i-1]&&b[i]!=c[i])
				{
					dp[i]=dp[i-1]*(c[i]-b[i]+1-num)%mod;
				}
				else if( b[i]<b[i-1]&&c[i-1]==c[i] || b[i]==b[i-1]&&c[i-1]<c[i] )
				{
						//dp[i] = dp[i-1]+1;
						dp[i]=dp[i-1];
				}
					num++; //已经选了多少种
			}
		  printf("%I64d\n",dp[n]);
		}
	}
    return 0;
}