// Original C++ solution (slower):


/**

    And in the end, the love you take is equal to the love you make.



        I LOVE YOU LAURA.

**/



#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/rope>



#define _CRT_SECURE_NO_DEPRECATE

#define _CRT_DISABLE_PERFCRIT_LOCKS



#define findOrder find_by_order

#define orderKey order_of_key



#define isOn(mask, i) (mask & (1 << i))

#define turn(mask, i) (mask |= (1 << i))

#define switch(mask, i) (mask &= ~(1 << i))

#define change(mask, i) (mask ^= (1 << i))

#define lowbit(mask) (mask & (-mask))

#define turnAll(mask, n) (mask = (1 << n) - 1)

#define switchLastBit(mask) (mask & (mask - 1))

#define turnLastZero(mask) (mask | (mask + 1))

#define modulo(x, mod) ((x) & (mod - 1))

#define log2(x) (31 - __builtin_clz(x))



#define mkp make_pair

#define pb push_back

#define pf push_front

#define eb emplace_back

#define ef emplace_front

#define ins insert

#define rsz resize

#define f first

#define s second

#define lb lower_bound

#define ub upper_bound



#define sz(x) ((int)x.size())

#define all(x) begin(x), end(x)

#define rall(x) rbegin(x), rend(x)



#define debug(x) (cerr << x << " ")



using namespace std;

using namespace __gnu_pbds;

using namespace __gnu_cxx;



typedef double db;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;



typedef pair<int, int> ii;

typedef pair<ll,ll> pll;

typedef pair<ld,ld> pdb;

typedef pair<int, ii> tri;

typedef complex<ld> cd;



typedef vector<int> vi;

typedef vector<ii> vii;

typedef vector<ld> vdb;

typedef vector<cd> vcd;

typedef vector<ll> vll;

typedef vector<string> vs;

typedef vector<vll> matrix;



typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_tree;



const ll mod = 1e9+7;

const int maxn = 405;

const int pi = 2 * acos(0.0);

const db eps = 1e-9;



ll dp[maxn][maxn], sum[maxn], arr[maxn];



ll solve_dp (int l, int r){

    if (l == r) return 0LL;



    ll &ans = dp[l][r];

    if (ans != -1) return ans;



    ans = (1LL << 60);

    for (int i = l; i < r; i++){

        ans = min(ans, solve_dp(l, i) + solve_dp(i + 1, r));

    }



    ans += (sum[r] - sum[l - 1]); // cost of combine element

    return ans;

}



int main()

{

    ios_base::sync_with_stdio(0);

    cin.tie(0); cout.tie(0);



    memset(dp, -1, sizeof dp);



    int n; cin >> n;



    for (int i = 1; i <= n; i++){

        cin >> arr[i];

        sum[i] = sum[i - 1] + arr[i];

    }



    cout << solve_dp(1, n) << "\n";



    return 0;

}


// Optimized solution to the same problem (faster):
#include <iostream>

#include <bits/stdc++.h>

using namespace std;



typedef vector<int> vi;

#define ll long long

#define PI 3.14159265359

#define x first

#define y second

#define point pair<int,int>

#define clr(x, val)	memset((x), (val), sizeof(x))



//you don't know math, unless you can do math!

ll dp[400][400];

ll OO = 1e15 + 10;



vector<ll> a, a_sum;

int main() {

    int n; cin>>n;

    a= vector<ll> (n);

    a_sum = vector<ll> (n);



    for(int i = 0; i < n; i++){

        cin>>a[i];

        a_sum[i] = a[i];

        if(i) a_sum[i]+= a_sum[i - 1];

    }





    for(int i = 0; i <  n - 1; i++){

        dp[i][i + 1] = a[i] + a[i + 1];

    }



    for(int sz = 2; sz < n; sz++){

        for(int i = 0; i < n - sz; i++){

         int j = i + sz;

         dp[i][j] = OO;



         for(int k  = i; k < j; k++)

            dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);



         dp[i][j] += a_sum[j];

         if(i - 1 >= 0) dp[i][j] -= a_sum[i - 1];

        }

    }



    cout<<dp[0][n - 1];

}


// Fastest known solution for this problem: