[思维]The Third Problem CF1699C

You are given a permutation a1,a2,…,an of integers from 0 to n−1. Your task is to find how many permutations b1,b2,…,bn are similar to permutation a.

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

MEX([al,al+1,…,ar])=MEX([bl,bl+1,…,br]),where the MEX of a collection of integers c1,c2,…,ck 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 nn is an array consisting of nn 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 tt (1≤t≤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 nn (1≤n≤10^5) — the size of permutation a.

The second line of each test case contains nn distinct integers a1,a2,…,an (0≤ai<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 aa, taken modulo 10^9+7.

Example

input

5
5
4 0 3 2 1
1
0
4
0 1 2 3
6
1 2 4 0 5 3
8
1 3 7 2 5 0 6 4

output

2
1
1
4
72

Note

For the first test case, the only permutations similar to a=[4,0,3,2,1] are [4,0,3,2,1] and [4,0,2,3,1].

For the second and third test cases, the given permutations are only similar to themselves.

For the fourth test case, there are 4 permutations similar to a=[1,2,4,0,5,3]:

• [1,2,4,0,5,3]
• [1,2,5,0,4,3]
• [1,4,2,0,5,3]
• [1,5,2,0,4,3]

题意： 给出一个长度为n的排列，询问有多少个排列和它相似，两排列相似定义为对于任意一个区间，两者的MEX均相等，MEX是区间内未出现的最小非负整数。

分析： 首先考虑一下0的位置是否可以改变，如果改变了，那0那个位置的区间MEX就会改变，所以0的位置是不能动的，再考虑1的位置是否可以改变，这时可以看一下0的位置和1的位置构成的区间，设为[l，r]，假如1改变了位置，那这个区间的MEX值一定是1，所以1的位置也不能变。接下来考虑2的位置，如果2在[l，r]之间，那2一定不能移动出这个区间，也就是2的位置只能选在这个区间内部，共有（区间长度-2）种选择，但如果2的位置不在[l，r]之间，那2的位置也就不能动了，这点可以分情况讨论得到，同时区间[l，r]也要被更新。对于剩下的数依此类推直到遍历完所有数，每个数的位置选择数相乘就得到了答案。

具体代码如下： 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
using namespace std;
 
int a[100005], pos[100005]; 
const int mod = 1e9+7;
 
signed main()
{
	int T;
	cin >> T;
	while(T--){
		int n, l, r;
		scanf("%d", &n); //n = 1
		for(int i = 1; i <= n; i++){	
			scanf("%d", &a[i]);
			if(a[i] == 0) l = i;
			if(a[i] == 1) r = i;
			pos[a[i]] = i;
		}
		if(r < l) swap(l, r);
		long long res = 1;
		for(int i = 2; i < n; i++){
			if(pos[i] > l && pos[i] < r) 
				res = res*(r-l+1-i)%mod;
			else{
				if(pos[i] < l)
					l = pos[i];
				if(pos[i] > r)
					r = pos[i];
			}
		}
		printf("%lld\n", res);
	} 
    return 0;
}