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>
#define pb emplace_back
#define fi first
#define se second
#define int long long
using namespace std;
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int l) { uniform_int_distribution<int> uid(0, l - 1); return uid(rang); }

int dp[500][500];

int dpp(int l, int r, std::vector<int>& pre) {
	if (dp[l][r] != -1) return dp[l][r];
	if (l >= r) return 0;
	dp[l][r] = 1e16;
	for (int i = l; i < r; ++i) dp[l][r] = min(dpp(l, i, pre) + dpp(i + 1, r, pre) + pre[r] - pre[l - 1], dp[l][r]);
	return dp[l][r];
}

int32_t main() {
	ios::sync_with_stdio(false); cin.tie(NULL); cout.precision(10); srand(chrono::high_resolution_clock::now().time_since_epoch().count());
	int n; cin >> n;
	memset(dp, -1, sizeof(dp));
	vector<int> a(n + 1); for (int o = 0; o < n; o++) { cin >> a[o + 1]; }
	std::vector<int> pre(n + 1, 0);
	for (int i = 1; i <= n; ++i)
		pre[i] = pre[i - 1] + a[i];
	int res = dpp(1, n, pre);
	cout << res;
}