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)

// in the name of Allah

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

#define endl '\n'
#define pb push_back
#define ll long long
#define double long double
#define sz(x) ((int) (x.size()))

#define fr first
#define se second

#define inf 1e9
#define max_n 123456
#define mod 1000000007

ll add(ll a, ll b) {
	a += b;
	if(a >= mod) return a - mod;
	return a;
}

ll mul(ll a, ll b) {
	a *= b;
	if(a >= mod) return a % mod;
	return a;
}

ll power(ll a, ll b) {
	ll p = 1;
	while(b) {
		if(b&1) p = mul(p, a);
		a = mul(a, a);
		b /= 2;
	}
	return p;
}

int n;
ll a[500];
ll sum[500];
ll dp[500][500];

ll memo(int L, int R) {
	if(L == R) return 0;
	if(L+1 == R) return a[L] + a[R];
	if(dp[L][R] != -1) return dp[L][R];
	ll res = 1e18;
	for(int i=L; i<R; i++) {
		res = min(res, sum[R] - sum[L-1] + memo(L, i) + memo(i+1, R));
	}
	return dp[L][R] = res;
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n;
	sum[0] = 0;
	for(int i=1; i<=n; i++) {
		cin >> a[i];
		sum[i] = sum[i-1] + a[i];
	}
	memset(dp, -1, sizeof dp);
	cout << memo(1, n) << endl;
	return 0;
}