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: Runtime Error)

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

#define nax 409
#define FOR(i,n,m) for(ll i =(n);i<(m);i++)
#define REP(i,n) FOR(i,0,n)
#define pb push_back

typedef unsigned long long ll;
typedef vector<ll> vi;

ll dp[nax][nax];
ll sum[nax];

ll f(ll low, ll top){
	if(low==top){
		return dp[low][top];
	}
	if(dp[low][top]!=-1LL){
		return dp[low][top];
	}
	dp[low][top] = f(low,low)+f(low+1,top)+sum[top];
	if(low>0) dp[low][top] -= sum[low-1];
	ll aux = sum[top];
	if(low>0) aux-= sum[low-1];
	FOR(i,low,top+1){
		dp[low][top] = min(f(low,i)+f(i+1,top)+aux,dp[low][top]);
	}
	return dp[low][top];
}

int main(){
	ll n; cin>>n;
	vi a(n);

	REP(i,n){
		cin>>a[i];
		sum[i] = a[i];
		if(i>0) sum[i] += sum[i-1];
	}

	memset(dp,-1LL,sizeof dp);
	REP(i,n){
		dp[i][i] = 0;
	}

	cout<<f(0,n-1);

	return 0;
}