// Slower C++ solution:


/**

    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;

}


// Faster C++ solution for the same problem (measured speedup: 2.313636118055861x):
/*input

3

1000000000 1000000000 1000000000

*/

#include<bits/stdc++.h>

#define fn(i,n) for(long long i=0;i<n;i++)

#define fn1(i,n) for(long long  i=1;i<=n;i++)

#define ll long long int

#define mp(x,y) make_pair(x,y)

using namespace std;

int main()

{

ios_base::sync_with_stdio(false);

cin.tie(nullptr);

ll n;cin>>n;

ll dp[n][n],a[n];

fn(i,n){cin>>a[i];dp[i][i]=0;}

fn(i,n-1)

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

fn1(i,n-1)a[i]+=a[i-1];

for(int l=2;l<n;l++)

{

	for(int k=0;k+l<n;k++)

	{

		if(k>0)

		dp[k][k+l]=a[k+l]-a[k-1];

		else

		dp[k][k+l]=a[k+l];

		ll p=1e18;

		fn(j,l)

		{

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

		}

		dp[k][k+l]+=p;

	}

}

//fn(i,n)

//cout<<a[i]<<" ";

/*

fn(i,n)

{

	fn(j,n)

	cout<<dp[i][j]<<" ";

	cout<<endl;

}*/

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

}