The Little Elephant likes permutations. This time he has a permutation A[1], A[2], ..., A[N] of numbers 1, 2, ...,N.

He calls a permutation A good, if the number of its inversions is equal to the number of its local inversions. The number of inversions is equal to the number of pairs of integers (i; j) such that 1 ≤ i < j ≤ N and A[i] > A[j], and the number of local inversions is the number of integers i such that 1 ≤ i < N and A[i] > A[i+1].

The Little Elephant has several such permutations. Help him to find for each permutation whether it is good or not. Print YES for a corresponding test case if it is good and NO otherwise.

Input

The first line of the input contains a single integer T, the number of test cases. T test cases follow. The first line of each test case contains a single integer N, the size of a permutation. The next line contains N space separated integers A[1], A[2], ..., A[N].

Output

For each test case output a single line containing the answer for the corresponding test case. It should beYES if the corresponding permutation is good and NO otherwise.

Constraints

1 ≤ T ≤ 474

1 ≤ N ≤ 100

It is guaranteed that the sequence A[1], A[2], ..., A[N] is a permutation of numbers 1, 2, ..., N.

Example

Input:41122 133 2 141 3 2 4Output:YESYESNOYES

总结一下有下面关系：

全局逆序数 == 起泡排序交换数据的交换次数

本题数据量小，能够使用暴力法计算。

可是这样时间效率是O(N*N)，要想减少到O(nlgn)那么就要使用merger sort。

以下一个类中暴力法和merge sort方法都包含了。

#pragma once#include 
     
      #include 
      
       class LittleElephantAndPermutations{	int localInverse(const int *A, const int n)	{		int ans = 0;		for (int i = 1; i < n; i++)		{			if (A[i-1] > A[i]) ans++;		}		return ans;	}	int globalInverse(const int *A, const int n)	{		int ans = 0;		for (int i = 0; i < n; i++)		{			for (int j = i+1; j < n; j++)			{				if (A[i] > A[j]) ans++;			}		}		return ans;	}	int *mergeArr(int *left, int L, int *right, int R, int &c)	{		int *lr = new int[L+R];		int i = 0, j = 0, k = 0;		while (i < L && j < R)		{			if (left[i] <= right[j])			{				lr[k++] = left[i++];			}			else			{				lr[k++] = right[j++];				c += L-i;			}		}		while (i < L)		{			lr[k++] = left[i++];		}		while (j < R)		{			lr[k++] = right[j++];		}		return lr;	}	int biGlobalInverse(int *&A, const int n)	{		if (n <= 1) return 0;		int mid = (n-1) >> 1;//记得n-1		int *left = new int[n];		int *right = new int[n];		int L = 0, R = 0;		for ( ; L <= mid; L++)		{			left[L] = A[L];		}		for (int i = mid+1; i < n; i++, R++)		{			right[R] = A[i];		}		delete [] A;		int c1 = biGlobalInverse(left, L);		int c2 = biGlobalInverse(right, R);		int c = 0;				A = mergeArr(left, L, right, R, c);		delete left;		delete right;		return c1+c2+c;	}public:	void run()	{		int T = 0, N = 0;		scanf("%d", &T);		while (T--)		{			scanf("%d", &N);			int *A = new int[N];			for (int i = 0; i < N; i++)			{				scanf("%d", &A[i]);			}			int loc = localInverse(A, N);			int glo = biGlobalInverse(A, N);			if (loc == glo) puts("YES");			else puts("NO");		}	}};int littleElephantAndPermutations(){	LittleElephantAndPermutations le;	le.run();	return 0;}