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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
long long MIN(long long a,long long b){return a<b?a:b;}
int main(){
  long long n,i,j,k,a[410],d[410][410]={};
  scanf("%lld",&n);
  for(i=0;i<410*410;i++)d[i/410][i%410]=1e18;
  for(i=0;i<n;i++)scanf("%lld",&a[i+1]);
  for(i=0;i<n;i++)a[i+1]+=a[i];
  for(i=0;i<n;i++)d[i][i+1]=a[i+1]-a[i];
  for(i=1;i<=n;i++){
    for(j=0;j+i<=n;j++){
      for(k=j;k<=j+i;k++)d[j][j+i]=MIN(d[j][j+i],d[j][k]+d[k][j+i]+a[j+i]-a[j]);
    }
  }
  printf("%lld\n",d[0][n]-a[n]);
  return 0;
}