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

using ll = long long; 
/*
dp[i][j] - minimum cost from i to j
*/

int main() {
    const int INF = 1e9+7;
    
    int n;
    
    scanf("%d", &n);
    
    vector<int> arr(n);
    
    vector<vector<ll>> dp(n, vector<ll>(n, -1));
    
    vector<vector<int>> mids(n, vector<int>(n));
    
    vector<ll> pref_sums(n+1);
    
    ll pref_sum = 0;
    
    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
        dp[i][i] = 0;
    }
    
    for (int i = 0; i <= n; i++) {
        pref_sum += arr[i];
        pref_sums[i] = pref_sum;
    }
    
    for (int length = 1; length < n; length++) {
        for (int left = 0; left+length < n; left++) {
            int right = left + length;
            //cout << left << " " << right << endl;
            for (int mid = left; mid < right; mid++) {
                ll left_cost = dp[left][mid];
                ll right_cost = dp[mid+1][right];
                ll range_sum = pref_sums[right] - (left-1 < 0 ? 0 : pref_sums[left-1]);
                ll my_cost = left_cost + right_cost + range_sum;
                //cout << total << endl;
                /*
                if (total < dp[left][right]) {
                    // update chosen mid & min value
                    dp[left][right] = my_cost;//total;
                    //mids[left][right] = mid;
                }
                */
                if (dp[left][right] == -1) {
                    dp[left][right] = my_cost;
                }
                else {
                    dp[left][right] = min(dp[left][right], my_cost);                
                }
            }
        }
    }
    /*
    vector<pair<int, int>> v; // (start, end)
    
    v.push_back({0, n-1});
    
    ll ans = 0; //-dp[0][n-1]; // the last one 
    
    while (!v.empty()) {
        pair<int, int> cur = v.back();
        v.pop_back();
        if (cur.second - cur.first + 1 >= 2) { // interval at least two
            ans += dp[cur.first][cur.second];
            int mid = mids[cur.first][cur.second];
            cout << "L: " << cur.first << " R: " << cur.second << " Total: " << dp[cur.first][cur.second] << " mid: " << mid << endl;
            v.push_back({cur.first, mid});
            v.push_back({mid+1, cur.second});    
        }
    }
    */
    
    
	/*
	for (int i = 0; i < n; i++) {
	    for (int j = 0; j < n; j++) {
	        for (int mid = i; mid <= j; mid++) {
	            ll cur = dp[i][mid];
	            if (mid+1 <= j) {
	                cur += dp[mid+1][j];
	            }
	            dp[i][j] = min(dp[i][j], cur);
	        }
	    }
	}
	*/
	
	//printf("%lld", ans);
	
	printf("%lld", dp[0][n-1]);
	
	return 0;
}