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)

/*input
3 3
abc


*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long, null_type, less<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
typedef long long int ll;
#define ld long double
#define pii pair<ll,ll>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define sz(_a) (ll)(_a.size())
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
const ll maxn=405;
const ll maxlg=__lg(maxn)+2;
const ll INF64=8000000000000000000LL;
const int INF=0x3f3f3f3f;
const ll MOD=ll(1e9+7);
const ld PI=acos(-1);
const ld eps=1e-9;

ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=res*a%MOD;
        a=a*a%MOD;
        b>>=1;
    }
    return res;
}
inline ll add(ll a,ll b){
	return ((a%MOD)+(b%MOD))%MOD;
}
inline ll sub(ll a,ll b){
	a -= b;
	if(a < 0)a += MOD;
	return a%MOD;
}
ll dp[maxn][maxn],pref[maxn],arr[maxn];
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin>>n;
    pref[0]=0;
    FILL(dp,INF);
    for(int i=1;i<=n;i++){
    	cin>>arr[i];
    	pref[i]=pref[i-1]+arr[i];
    	dp[i][i]=0;
    }
    
    for(int i=n;i>=1;i--){
    	for(int j=i+1;j<=n;j++){
    		for(int k=i;k<j;k++){
    			dp[i][j]=min(dp[i][k]+dp[k+1][j]+pref[j]-pref[i-1],dp[i][j]);
    		}
    	}
    }
    cout<<dp[1][n];
}