#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef long long ll;
typedef long double lld;
typedef long long int lli;
using namespace std;
const int N = 1000001;
const int MOD=1e9+7;
const bool DEBUG = 1;
#define sd(x) scanf("%d", &x)
#define sd2(x, y) scanf("%d%d", &x, &y)
#define sd3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define endl "\n"
#define fi first
#define se second
#define eb emplace_back
#define fbo find_by_order
#define ook order_of_key
#define pb(x) push_back(x)
#define mp(x, y) make_pair(x, y)
#define LET(x, a) __typeof(a) x(a)
#define foreach(it, v) for (LET(it, v.begin()); it != v.end(); it++)
#define MEMS(a, b) memset(a, b, sizeof(a))
#define _			\
    ios_base::sync_with_stdio(false);			\
    cin.tie(NULL);			\
    cout.tie(NULL);
#define __ 			\
    freopen("input.txt", "r", stdin);			\
    freopen("output.txt", "w", stdout);
#define all(c) c.begin(), c.end()
#define inf 1000000000000000001
#define epsilon 1e-6
#define int ll
#define RUN_T			 \
    int _t; 			 \
    cin >> _t;			 \
    while (_t--)			 
#define tr(...)				\
if (DEBUG) {				\
cout << __FUNCTION__ << ' ' << __LINE__ << " = ";				\
trace(#__VA_ARGS__, __VA_ARGS__);				\
}
template <typename S, typename T>
ostream &operator<<(ostream &out, pair<S, T> const &p) {
    out << '(' << p.fi << ", " << p.se << ')';
    return out;
}
template <typename T>
ostream &operator<<(ostream &out, set<T> const &v) {
    for (auto i = v.begin(); i != v.end(); i++)
        out << (*i) << ' ';
    return out;
}
template <typename T, typename V>
ostream &operator<<(ostream &out, map<T, V> const &v) {
    for (auto i = v.begin(); i != v.end(); i++)
        out << "\n" << (i->first) <<  ":"  << (i->second);
    return out;
}
template <typename T, typename V>
ostream &operator<<(ostream &out, unordered_map<T, V> const &v) {
    for (auto i = v.begin(); i != v.end(); i++)
        out << "\n" << (i->first) <<  ":"  << (i->second);
    return out;
}
template <typename T>
ostream &operator<<(ostream &out, multiset<T> const &v) {
    for (auto i = v.begin(); i != v.end(); i++)
        out << (*i) << ' ';
    return out;
}
template <typename T>
ostream &operator<<(ostream &out, unordered_set<T> const &v) {
    for (auto i = v.begin(); i != v.end(); i++)
        out << (*i) << ' ';
    return out;
}
template <typename T>
ostream &operator<<(ostream &out, unordered_multiset<T> const &v) {
    for (auto i = v.begin(); i != v.end(); i++)
        out << (*i) << ' ';
    return out;
}
template <typename T> ostream &operator<<(ostream &out, vector<T> const &v) {
    ll l = v.size();
    for (ll i = 0; i < l - 1; i++)
        out << v[i] << ' ';
    if (l > 0)
        out << v[l - 1];
    return out;
}
template <typename T> void trace(const char *name, T &&arg1) {
    cout << name <<  ":"  << arg1 << endl;
}
template <typename T, typename... Args>
void trace(const char *names, T &&arg1, Args &&... args) {
    const char *comma = strchr(names + 1, ',');
    cout.write(names, comma - names) <<  ":"  << arg1 <<  "|" ;
    trace(comma + 1, args...);
}
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
int gcd(int a, int b) {
     if (b == 0)
        return a;
    return gcd(b, a % b);
}
int32_t main() {
    _
    int n;
    cin >> n;
    vector<int>v(n + 1),s(n+1,0);
    vector<vector<int>>dp(n + 1, vector<int>(n+1,0));
    for(int i = 1; i <= n; i++){
         cin >> v[i];
         s[i] = s[i-1] + v[i];
         dp[i][i] = 0;
    }
    int ans = 0;
    for(int len = 2; len <= n ; len++){
        for(int l = 1; l + len - 1 <= n; l++){
            int r = l + len - 1;
            dp[l][r] = inf;
            for(int k = l; k < r; k++){
                dp[l][r] = min( dp[l][r] , dp[l][k] + dp[k+1][r] + s[r] - s[l-1]);
            }
        }
    }
    cout << dp[1][n] << endl;
    return 0;
}
// 3 2 2 2
// my previous greedy merging techinque will fail in this test case as according to my algo answer will be 
// (3+(2+2)+2) but optimal is ((3+2) + (2+2))

//in merging technique start form the back
// dp[l][r] — minimum cost to merge from l to r Notice that in the last operation we will merge two 
// consecutive blocks, and their total sum will be the sum of all slimes.
//  So dp[l][r] = min(dp[l][k] + dp[k][r]) + a[l] + a[l + 1] + ... + a[r]. Which is O(n3)

Measured memory_bytes: 1536