Problem:
Score : 100 points

Problem Statement

There are N slimes lining up in a row.
Initially, the i-th slime from the left has a size of a_i.

Taro is trying to combine all the slimes into a larger slime.
He will perform the following operation repeatedly until there is only one slime:

Choose two adjacent slimes, and combine them into a new slime. The new slime has a size of x + y, where x and y are the sizes of the slimes before combining them. Here, a cost of x + y is incurred. The positional relationship of the slimes does not change while combining slimes.

Find the minimum possible total cost incurred.

Constraints

All values in input are integers.

2 \leq N \leq 400

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_N

Output

Print the minimum possible total cost incurred.

Sample Input 1

4
10 20 30 40

Sample Output 1

190

Taro should do as follows (slimes being combined are shown in bold):

(10, 20, 30, 40) → (30, 30, 40)

(30, 30, 40) → (60, 40)

(60, 40) → (100)

Sample Input 2

5
10 10 10 10 10

Sample Output 2

120

Taro should do, for example, as follows:

(10, 10, 10, 10, 10) → (20, 10, 10, 10)

(20, 10, 10, 10) → (20, 20, 10)

(20, 20, 10) → (20, 30)

(20, 30) → (50)

Sample Input 3

3
1000000000 1000000000 1000000000

Sample Output 3

5000000000

The answer may not fit into a 32-bit integer type.

Sample Input 4

6
7 6 8 6 1 1

Sample Output 4

68

Taro should do, for example, as follows:

(7, 6, 8, 6, 1, 1) → (7, 6, 8, 6, 2)

(7, 6, 8, 6, 2) → (7, 6, 8, 8)

(7, 6, 8, 8) → (13, 8, 8)

(13, 8, 8) → (13, 16)

(13, 16) → (29)

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<string>
#include<deque>
#include<unordered_map>
#include<unordered_set>
#include<queue>
#include<algorithm>
#include<cmath>
#include<numeric>
using namespace std;

typedef long long LL;
const int MOD = 1e9+7;

void solve(){
	int n;
	cin>>n;
	vector<int> nums(n);
	for(int i=0;i<n;i++) cin>>nums[i];
	LL nil = 1e15+7;
	vector<vector<LL>> dp(n,vector<LL>(n,nil));
	vector<LL> sif(n+1,0ll);
	for(int i=0;i<n;i++){
		sif[i+1] = sif[i]+nums[i];
		dp[i][i] = 0;
	}

	for(int len=2;len<=n;len++){
		for(int i=0;i<(n-len+1);i++){
			int j = i+len-1;
			for(int k=i;k<j;k++){
				dp[i][j] = min(dp[i][k]+sif[j+1]-sif[i]+dp[k+1][j],dp[i][j]);
			}
		}
	}
	cout<<dp[0][n-1]<<endl;
}


int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int T = 1;
	// cin>>T;
	while(T--){
		solve();
	}
	return 0;
}