B. Kalindrome Array

Problem - B - Codeforces

B. Kalindrome Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

An array [b1,b2,…,bm][b1,b2,…,bm] is a palindrome, if bi=bm+1−ibi=bm+1−i for each ii from 11 to mm. Empty array is also a palindrome.

An array is called kalindrome, if the following condition holds:

• It's possible to select some integer xx and delete some of the elements of the array equal to xx, so that the remaining array (after gluing together the remaining parts) is a palindrome.

Note that you don't have to delete all elements equal to xx, and you don't have to delete at least one element equal to xx.

For example :

• [1,2,1][1,2,1] is kalindrome because you can simply not delete a single element.
• [3,1,2,3,1][3,1,2,3,1] is kalindrome because you can choose x=3x=3 and delete both elements equal to 33, obtaining array [1,2,1][1,2,1], which is a palindrome.
• [1,2,3][1,2,3] is not kalindrome.

You are given an array [a1,a2,…,an][a1,a2,…,an]. Determine if aa is kalindrome or not.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤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 (1≤n≤2⋅1051≤n≤2⋅105) — the length of the array.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n) — elements of the array.

It's guaranteed that the sum of nn over all test cases won't exceed 2⋅1052⋅105.

Output

For each test case, print YES if aa is kalindrome and NO otherwise. You can print each letter in any case.

Example

input

Copy

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

output

Copy

YES
YES
NO
YES

Note

In the first test case, array [1][1] is already a palindrome, so it's a kalindrome as well.

In the second test case, we can choose x=2x=2, delete the second element, and obtain array [1][1], which is a palindrome.

In the third test case, it's impossible to obtain a palindrome.

In the fourth test case, you can choose x=4x=4 and delete the fifth element, obtaining [1,4,4,1][1,4,4,1]. You also can choose x=1x=1, delete the first and the fourth elements, and obtain [4,4,4][4,4,4].

双指针，左右相等继续推进，一旦不相等，就意味着必须删掉一个，而且答案就在这两者之间，所以枚举两者删除情况即可。题目忽悠人，说删几个x都行，其实必须都删才最好，因为如果两个x能左右对称那么删与不删一个样，一旦不对称还不如都删了。

#include <iostream>
# include<algorithm>
# include<cstring>

using namespace std;

typedef long long int ll;

int book[200000+10];
int b[200000+10];
int a[200000+10];
int main()
{
     int t;

     cin>>t;

     while(t--)
     {
         int n;

         cin>>n;

         memset(book,0,sizeof(book));

         for(int i=1;i<=n;i++)
         {
             cin>>a[i];

         }

         if(n==1||n==2)
         {
             cout<<"YES"<<'\n';

             continue;
         }
          int l=1,r=n;

          int ans=0;
          while(l<r)
          {

              if(a[l]==a[r])
              {
                  l++;

                  r--;

              }
              else
              {
                  int len=0;

                  for(int i=1;i<=n;i++)
                  {
                      if(a[i]!=a[l])
                      {
                          len++;
                          b[len]=a[i];

                      }
                  }
                  int flag=0;
                  for(int i=1;i<=len/2;i++)
                  {
                      if(b[i]!=b[len-i+1])
                      {
                          flag=1;
                      }
                  }

                  if(!flag)
                  {
                      ans=1;

                      break;
                  }






                   len=0;

                  for(int i=1;i<=n;i++)
                  {
                      if(a[i]!=a[r])
                      {
                          len++;
                          b[len]=a[i];

                      }
                  }
                   flag=0;
                  for(int i=1;i<=len/2;i++)
                  {
                      if(b[i]!=b[len-i+1])
                      {
                          flag=1;
                      }
                  }

                  if(!flag)
                  {
                      ans=1;

                      break;
                  }

                  break;
              }
          }


          if(ans||l>=r)
          {
              cout<<"YES"<<endl;
          }
          else
          {
              cout<<"NO"<<endl;
          }
     }

    return 0;
}