Codeforces Round #336 (Div. 1)B. Zuma （区间DP）

B. 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 ci. 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 ci (1 ≤ ci ≤ 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.
题意：给一个长度为n的序列，每次可以删除任意长度的的回文串，问最少操作数。
题解：区间dp
dp[i][j]表示消除区间[I，j]的最小操作数。
当a[l]==a[r]时，dp[l][r]=solve(l+1,r-1).
否则 枚举断点。
代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
using namespace std;
int n;
int a[505];
int dp[505][505];
int  solve(int l,int r)
{
    if(l>=r) return 1;
    if(dp[l][r]!=-1) return dp[l][r];
    dp[l][r]=1e9;
    if(a[l]==a[r]) dp[l][r]=solve(l+1,r-1);
    for(int i=l; i1,r));
        return dp[l][r];
}
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    memset(dp,-1,sizeof(dp));
    printf("%d\n",solve(1,n));
    return 0;
}