CodeForces 607B zuma

Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color c_i. The goal of the game is to destroy all the gemstones in the line as quickly as possible.

In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?

Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.

Input

The first line of input contains a single integer n (1 ≤ n ≤ 500) — the number of gemstones.

The second line contains n space-separated integers, the i-th of which is c_i (1 ≤ c_i ≤ n) — the color of the i-th gemstone in a line.

Output

Print a single integer — the minimum number of seconds needed to destroy the entire line.

Examples

Input

3
1 2 1

Output

1

Input

3
1 2 3

Output

3

Input

7
1 4 4 2 3 2 1

Output

2

Note

In the first sample, Genos can destroy the entire line in one second.

In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.

In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.

题解：题解：区间DP。若区间i~j为回文串，那么i+1~j-1也必然为回文串，dp[i][j]=dp[i+1][j-1],

如果不是，这就枚举中间点，区间DP裸题；

参考代码为：

 #include<bits/stdc++.h>
 using namespace std;

 int num[], dp[][];
 int main()
 {
     int n;
     while (~scanf("%d", &n))
     {
         memset(num, , sizeof(num));
         memset(dp,0x3f,sizeof(dp));
         for (int i = ; i <= n; i++)
             scanf("%d", &num[i]);
         for (int i = ; i <= n; i++)
             dp[i][i] = ;
         for (int i = n; i >= ; i--)
             for (int j = i + ; j <= n; j++)
             {
                 if (num[i] == num[j])
                 {
                     if (abs(j - i) != )
                         dp[i][j] = dp[i + ][j - ];
                     else
                         dp[i][j] = ;
                 }

                 for (int k = i; k<j; k++)
                 {
                     dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + ][j]);
                 }
             }
         printf("%d\n", dp[][n]);
     }
     return ;
 }