We are all familiar with sorting algorithms: quick sort, merge sort, heap sort, insertion sort, selection sort, bubble sort, etc. But sometimes it is an overkill to use these algorithms for an almost sorted array.  We say an array is sorted if its elements are in non-decreasing order or non-increasing order. We say an array is almost sorted if we can remove exactly one element from it, and the remaining array is sorted. Now you are given an array a1,a2,…,ana1,a2,…,an, is it almost sorted?

Input

The first line contains an integer TT indicating the total number of test cases. Each test case starts with an integer nn in one line, then one line with nn integers a1,a2,…,ana1,a2,…,an.  1≤T≤20001≤T≤2000  2≤n≤1052≤n≤105  1≤ai≤1051≤ai≤105  There are at most 20 test cases with n>1000n>1000.

Output

For each test case, please output "`YES`" if it is almost sorted. Otherwise, output "`NO`" (both without quotes).

Sample Input

3
3
2 1 7
3
3 2 1
5
3 1 4 1 5

Sample Output

YES
YES
NO

題意：給你個序列 讓你判斷去掉一個數，該數列是否是一個非遞增序列或者是非遞減序列。

思路：分別求最長上升子序列和最長下降子序列的長度 ，如果長度為n-1 或 n 則YES 否則NO（複雜度O(nlogn)）

#include<bits/stdc++.h>
#define CaseT int t;scanf("%d",&t);while(t--)
#define ms(a,x) memset(a,x,sizeof(a))
using namespace std;

#define N 107
#define MAX 100002

typedef long long ll;

int n,m;
int a[MAX];
int d[MAX];
int u[MAX];
int b[MAX];

void solve(){
	int n;
	cin>>n;
	ms(a,0);ms(b,0);ms(d,0);ms(u,0);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		b[n-i+1]=a[i];
	}
	
	u[1]=a[1];
	d[1]=b[1];
	int up=1,down=1;
	for(int i=2;i<=n;i++){
		if(a[i]>=u[up])//非遞減 就加上等號。
			u[++up]=a[i];
		else {
			int pos=upper_bound(u+1,u+1+up,a[i])-u;
			u[pos]=a[i];
		}
		
		if(b[i]>=d[down])
			d[++down]=b[i];
		else {
			int pos=upper_bound(d+1,d+1+down,b[i])-d;
			d[pos]=b[i];
		}
	}
	
	if(up==n||up==n-1||down==n||down==n-1)
		cout<<"YES\n";
	else cout<<"NO\n";
	
}
 
int main(){
	CaseT
	solve();
	return 0;
}