CodeForces - 1312E Array Shrinking 区间dp

You are given an array a1,a2,…,ana1,a2,…,an. You can perform the following operation any number of times:

Choose a pair of two neighboring equal elements ai=ai+1ai=ai+1 (if there is at least one such pair).
Replace them by one element with value ai+1ai+1.

After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array aa you can get?

Input

The first line contains the single integer nn (1≤n≤5001≤n≤500) — the initial length of the array aa.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤10001≤ai≤1000) — the initial array aa.

Output

Print the only integer — the minimum possible length you can get after performing the operation described above any number of times.

Examples

Input

5
4 3 2 2 3

Output

Input

7
3 3 4 4 4 3 3

Output

Input

3
1 3 5

Output

Input

1
1000

Output

Note

In the first test, this is one of the optimal sequences of operations: 44 33 22 22 33 →→ 44 33 33 33 →→ 44 44 33 →→ 55 33.

In the second test, this is one of the optimal sequences of operations: 33 33 44 44 44 33 33 →→ 44 44 44 44 33 33 →→ 44 44 44 44 44 →→ 55 44 44 44 →→ 55 55 44 →→ 66 44.

In the third and fourth tests, you can't perform the operation at all.

区间dp

dp数组存l到r里面的个数

当相邻dp数组个数唯一且相等时更新

#include<bits/stdc++.h>
const int maxn = 1e3 + 5;
int num[maxn][maxn];
int dp[maxn][maxn];
using namespace std;int main(){int n;cin >> n;for(int i = 1; i <= n; i++){cin >> num[i][i];dp[i][i] = 1;}for(int i = 1; i <= n; i++){for(int j = i; j <= n; j++){dp[i][j] = j - i + 1;}}for(int len = 2; len <= n; len++){for(int l = 1; l + len - 1 <= n; l++){int r = l + len - 1;for(int k = l; k < r; k++){dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r]);if(dp[l][k] == 1 && dp[k + 1][r] == 1 && num[l][k] == num[k + 1][r]){dp[l][r] = 1;num[l][r] = num[l][k] + 1;                  }}}}cout << dp[1][n] << endl;
}

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