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;

#define fto(i, s, e) for(int i = s; i <= e; ++i)
#define fto1(i, s, e) for(int i = s; i < e; ++i)
#define fdto(i, s, e) for(int i = s; i >= e; --i)
#define fit(var, it) for(__typeof(var.begin()) it = var.begin(); it != var.end(); ++it)
#define frit(var, it) for(__typeof(var.rbegin()) it = var.rbegin(); it != var.rend(); ++it)

#define newl '\n'
#define debugt cerr << 0.001*clock() << newl
#define debug1(x, i) cout << #x << '[' << i << "] = " << x[i] << newl
#define debug(v, l, r) fto(_i, l, r) debug1(v, _i)
#define debug2(x, i, j) cout << #x << '[' << i << "][" << j << "] = " << x[i][j] << newl
#define debug3(x, i, j, k) cout << #x << '[' << i << "][" << j << "][" << k << "] = " << x[i][j][k] << newl

#define ll long long
#define ii pair<int, int>
#define ff first
#define ss second
#define pb push_back
#define glen(v) int((v).size())

template<typename T1, typename T2> ostream& operator<< (ostream &os, pair<T1, T2> const &v) {
	return os << '(' << v.ff << ", " << v.ss << ')';
}

#define eps 1e-15
#define oo 1000000007
#define OO 1000000000000000003LL
#define maxn 403

int n;
int a[maxn];
ll sum[maxn];
ll f[maxn][maxn];

int main() {
	//ios_base::sync_with_stdio(0); cin.tie(0);

	scanf("%d", &n);

	fto (i, 1, n) {
		scanf("%d", &a[i]);
		sum[i] = sum[i-1] + a[i];
	}

	fdto (i, n, 1) {
		fto (j, i+1, n) {
			f[i][j] = OO;
			fto1 (k, i, j) {
				f[i][j] = min(f[i][j], f[i][k] + f[k+1][j] + sum[j] - sum[i-1]);
			}
		}
	}

	printf("%lld\n", f[1][n]);

	return 0;
}