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 ll          long long int
#define pb          push_back
#define pll         pair<ll,ll>
#define speed_up    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mode        998244353
#define mod         1000000007
#define inf         1000000000000000000LL
#define vl          vector<ll>
#define F           first
#define S           second
#define db          long double
#define sz(x)       (ll)x.size()
#define fix(n)      cout<<fixed<<setprecision(n)
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define all(x)      x.begin(),x.end()
#define mset(x)     memset(x,0,sizeof x)
#define pi          3.14159265358979323

ll dp[404][404], a[404], sum[404];

ll f(ll l, ll r)
{
       if(l==r) return 0;
       ll &ans = dp[l][r];
       if(ans!=-1) return ans;
       ans = inf;
       ll s = sum[r] - sum[l-1];
       rep(i, l, r)
              ans = min(ans, f(l, i)+f(i+1, r)+ s);
       return ans;
}

void solve()
{
       ll n;
       cin>>n;
       rep(i,1,n+1) cin>>a[i];
       sum[0]=0;
       rep(i,1,n+1) sum[i] = sum[i-1]+a[i];
       memset(dp, -1, sizeof dp);
       cout<<f(1, n);
}

int main() {
	ll T = 1;
	speed_up
	//cin>>T;
	while(T--)
	    solve();
	return 0;
}