The Third Problem

You are given a permutation $a_{1}, a_{2}, \dots, a_{n}$ of integers from $0$ to $n - 1$ . Your task is to find how many permutations $b_{1}, b_{2}, \dots, b_{n}$ are similar to permutation $a$ .

Two permutations $a$ and $b$ of size $n$ are considered similar if for all intervals $[l, r]$ ( $1 \leq l \leq r \leq n$ ), the following condition is satisfied:

MEX ([a l, a l + 1, \dots, a r]) = MEX ([b l, b l + 1, \dots, b r]),

where the $MEX$ of a collection of integers $c_{1}, c_{2}, \dots, c_{k}$ is defined as the smallest non-negative integer $x$ which does not occur in collection $c$ . For example, $MEX ([1, 2, 3, 4, 5]) = 0$ , and $MEX ([0, 1, 2, 4, 5]) = 3$ .

Since the total number of such permutations can be very large, you will have to print its remainder modulo $10^{9} + 7$ .

In this problem, a permutation of size $n$ is an array consisting of $n$ distinct integers from $0$ to $n - 1$ in arbitrary order. For example, $[1, 0, 2, 4, 3]$ is a permutation, while $[0, 1, 1]$ is not, since $1$ appears twice in the array. $[0, 1, 3]$ is also not a permutation, since $n = 3$ and there is a $3$ in the array.

Input

Each test contains multiple test cases. The first line of input contains one integer $t$ ( $1 \leq t \leq 10^{4}$ ) — the number of test cases. The following lines contain the descriptions of the test cases.

The first line of each test case contains a single integer $n$ ( $1 \leq n \leq 10^{5}$ ) — the size of permutation $a$ .

The second line of each test case contains $n$ distinct integers $a_{1}, a_{2}, \dots, a_{n}$ ( $0 \leq a_{i} < n$ ) — the elements of permutation $a$ .

It is guaranteed that the sum of $n$ across all test cases does not exceed $10^{5}$ .

Output

For each test case, print a single integer, the number of permutations similar to permutation $a$ , taken modulo $10^{9} + 7$ .

题目大意：给出一个0到n-1的排列，问有多少个0到n-1的全排列满足与原序列对应的任意字串的mex值相等，mex值为区间内未出现的最小非负整数。

思路：容易观察发现，0和1的位置是固定的，假设1在0右边，则有三种情况：1. 2在1右边，2.2在0左边，3.2在1与2中间。可以发现，因为0到1的mex是固定的，所以情况一与情况二的2便固定，考虑3的时候便要3在考虑0到2或者2到1（指的是原序列的地址）的情况，而情况三2便可以放在0到1之间任意空位。依次更新即可。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;

#define int long long
const int maxn=1e5+5,mod=1e9+7;
int t,n,a[maxn];

void solve() {
	cin>>n;
	for(int i=1;i<=n;i++) {
		int x;cin>>x;
		a[x]=i;
	}
	int l=a[0],r=a[0];
	int ans=1;
	for(int i=1;i<n;i++) {
		if(a[i]>r) r=a[i];
		else if(a[i]<l) l=a[i];
		else ans= (ans*(r-l-i+1))%mod;
	}
	cout<<ans<<'\n';
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	cin>>t;
	while(t--) {
		solve();
	}
}