D. Zuma time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output

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.

題目大意：給你一個長度為n的序列，每一次可以消去一段迴文串，問你最少消去多少次，能夠消除整個序列。

思路：

1、設定dp【i】【j】表示消去從i到j這段序列需要用的最小次數。

2、那麼不難理解：

dp【i】【j】=min（dp【i】【k】+dp【k】【j】，dp【i】【j】）；

if（a【i】==a【l】）dp【i】【j】=min（dp【i】【j】，dp【i+1】【l-1】）；

3、加以深搜，記憶化dp即可。

Ac程式碼：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int dp[805][805];
int a[805];
int n;
int Dfs(int l,int r)
{
    if(dp[l][r]==0x3f3f3f3f)
    {
        if(l==r)return 1;
        if(l+1==r)
        {
            if(a[l]==a[r])return 1;
            else return 2;
        }
        if(a[l]==a[r])
        {
            dp[l][r]=min(dp[l][r],Dfs(l+1,r-1));
        }
        for(int i=l;i<=r;i++)
        {
            dp[l][r]=min(Dfs(l,i)+Dfs(i+1,r),dp[l][r]);
        }
        return dp[l][r];
    }
    else return dp[l][r];
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(dp,0x3f3f3f3f,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        dp[1][n]=Dfs(1,n);
        printf("%d\n",dp[1][n]);
    }
}