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;

template <class T> inline bool chmin(T &a, T b) { /* {{{ */
  if (a > b) {
    a = b;
    return true;
  }
  return false;
} /* }}} */

int main() {
  int N;
  cin >> N;
  vector<long long> A(N);
  for (auto &a : A) cin >> a;

  vector<long long> AS(N + 1);
  for (int i = 0; i < N; i++) {
    AS[i + 1] = AS[i] + A[i];
  }

  vector<vector<long long>> dp(N + 5, vector<long long>(N + 5, 1e15));
  for (int l = N; l >= 0; l--) {
    for (int r = l; r <= N; r++) {
      if (l == r) {
        dp[l][r] = 0;
        continue;
      }
      for (int c = l; c < r; c++) {
        chmin(dp[l][r], dp[l][c] + dp[c + 1][r] + (AS[r + 1] - AS[l]));
      }
    }
  }
  cout << dp[0][N - 1] << endl;

  return 0;
}
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