#include <bits/stdc++.h>

using namespace std;

const int inf = 0x3f3f3f3f;
const int N = 2005; 

int T, n;
int a[N], dp[N][N];

int main() {
	cin >> T;
	while(T--) {
		scanf("%d", &n);
		for(int i = 1; i <= n; i++) scanf("%d", a + i);
		memset(dp, 0x3f, sizeof dp);
		dp[1][1] = -inf;
		for(int i = 1; i < n; i++) {//由于我的转移是转移i+1，所以i<n。
			for(int j = 1; j <= min(n / 2, i); j++) {//剪去无用状态
				if(a[i] < a[i + 1]) dp[i + 1][j + 1] = min(dp[i][j], dp[i + 1][j + 1]);
				if(a[i + 1] > dp[i][j]) dp[i + 1][i - j + 1] = min(a[i], dp[i + 1][i - j + 1]);//转移
			}
		}
		if(dp[n][n / 2] == inf) puts("No!");
		else puts("Yes!");
	}	
	return 0;
}