C. Element Extermination

C. Element Extermination

C. Element Extermination

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array aa of length nn, which initially is a permutation of numbers from 11 to nn. In one operation, you can choose an index ii (1≤i

For example, if you have the array [1,3,2][1,3,2], you can choose i=1i=1 (since a1=1

Is it possible to make the length of this array equal to 11 with these operations?

Input

The first line contains a single integer tt (1≤t≤2⋅1041≤t≤2⋅104)  — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer nn (2≤n≤3⋅1052≤n≤3⋅105)  — the length of the array.

The second line of each test case contains nn integers a1a1, a2a2, ..., anan (1≤ai≤n1≤ai≤n, aiai are pairwise distinct) — elements of the array.

It is guaranteed that the sum of nn over all test cases doesn't exceed 3⋅1053⋅105.

Output

For each test case, output on a single line the word "YES" if it is possible to reduce the array to a single element using the aforementioned operation, or "NO" if it is impossible to do so.

Example

input

Copy

4
3
1 2 3
4
3 1 2 4
3
2 3 1
6
2 4 6 1 3 5

output

Copy

YES
YES
NO
YES

Note

For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):

[1,2,3]→[1,2]→[1][1,2,3]→[1,2]→[1]

[3,1,2,4]→[3,1,4]→[3,4]→[4][3,1,2,4]→[3,1,4]→[3,4]→[4]

[2,4,6,1,3,5]→[4,6,1,3,5]→[4,1,3,5]→[4,1,5]→[4,5]→[4][2,4,6,1,3,5]→[4,6,1,3,5]→[4,1,3,5]→[4,1,5]→[4,5]→[4]

=========================================================================

这种问题，下下策是用一天时间妄图用算法高级数据结构去解，下策是用半天模拟，中策是半小时证明，上策是写4的全部排列五分钟找出来规律

# include
# include
# include
 
using namespace std;
 
 
int main ()
{
 
     int t;
     cin>>t;
 
     while(t--)
     {
         int n;
         cin>>n;
         int x,y;
         cin>>x;
         for(int i=1;i
         {
             cin>>y;
         }
         if(x>y)
         {
             cout<<"NO"<
         }
         else
            cout<<"YES"<
     }
 
 
    return  0;
 
}