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)

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize(3)
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC target("sse3","sse2","sse")
//#pragma GCC target("avx","sse4","sse4.1","sse4.2","ssse3")
//#pragma GCC target("f16c")
//#pragma GCC optimize("inline","fast-math","unroll-loops","no-stack-protector")
//#pragma GCC diagnostic error "-fwhole-program"
//#pragma GCC diagnostic error "-fcse-skip-blocks"
//#pragma GCC diagnostic error "-funsafe-loop-optimizations"
//#pragma GCC diagnostic error "-std=c++14"
#include "bits/stdc++.h"
#include "ext/pb_ds/tree_policy.hpp"
#include "ext/pb_ds/assoc_container.hpp"

#define PB push_back
#define PF push_front
#define LB lower_bound
#define UB upper_bound
#define fr(x) freopen(x,"r",stdin)
#define fw(x) freopen(x,"w",stdout)
#define iout(x) printf("%d\n",x)
#define lout(x) printf("%lld\n",x)
#define REP(x, l, u) for(ll x = l;x<u;x++)
#define RREP(x, l, u) for(ll x = l;x>=u;x--)
#define complete_unique(a) a.erase(unique(a.begin(),a.end()),a.end())
#define mst(x, a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define MP make_pair
#define lowbit(x) ((x)&(-(x)))
#define lson (ind<<1)
#define rson (ind<<1|1)
#define se second
#define fi first
#define sz(x) ((int)x.size())
#define EX0 exit(0);

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
using namespace __gnu_pbds; //required
using namespace std;
template<typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef vector<ll> VLL;
typedef vector<int> VI;
const int block_size = 320;
typedef complex<ll> point;
const ll mod = 1e9 + 7;
const ll inf = 1e9 + 7;
const ld eps = 1e-9;
const db PI = atan(1) * 4;

template<typename T>
inline int sign(const T &a) {
    if(a < 0)return -1;
    if(a > 0)return 1;
    return 0;
}

string to_string(string s) { return '"' + s + '"'; }

string to_string(const char *s) { return to_string((string) s); }

string to_string(bool b) { return (b ? "true" : "false"); }

template<typename A, typename B>
string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }

template<typename A>
string to_string(A v) {
    bool first = true;
    string res = "{";
    for(const auto &x : v) {
        if(!first) { res += ", "; }
        first = false;
        res += to_string(x);
    }
    res += "}";
    return res;
}

void debug_out() { cerr << endl; }

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
}

#ifndef ONLINE_JUDGE
#define dbg(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define dbg(...) {}
#endif

template<typename T, typename S>
inline bool upmin(T &a, const S &b) { return a > b ? a = b, 1 : 0; }

template<typename T, typename S>
inline bool upmax(T &a, const S &b) { return a < b ? a = b, 1 : 0; }

template<typename T>
inline void in(T &x) {
    x = 0;
    T f = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while(isdigit(ch)) {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    x *= f;
}

ull twop(ll x) { return 1ULL << x; }

ll MOD(ll a, ll m) {
    a %= m;
    if(a < 0)a += m;
    return a;
}

ll inverse(ll a, ll m) {
    a = MOD(a, m);
    if(a <= 1)return a;
    return MOD((1 - inverse(m, a) * m) / a, m);
}

template<typename A, typename B>
inline void in(A &x, B &y) {
    in(x);
    in(y);
}

template<typename A, typename B, typename C>
inline void in(A &x, B &y, C &z) {
    in(x);
    in(y);
    in(z);
}

template<typename A, typename B, typename C, typename D>
inline void in(A &x, B &y, C &z, D &d) {
    in(x);
    in(y);
    in(z);
    in(d);
}

template<typename T>
T sqr(T x) { return x * x; }

ll gcd(ll a, ll b) {
    while(b != 0) {
        a %= b;
        swap(a, b);
    }
    return abs(a);
}

ll fast(ll a, ll b, ll mod) {
    if(b < 0)a = inverse(a, mod), b = -b;
    ll ans = 1;
    while(b) {
        if(b & 1) {
            b--;
            ans = ans * a % mod;
        } else {
            a = a * a % mod;
            b /= 2;
        }
    }
    return ans % mod;
}


namespace SOLVE {
    const int maxn = 410;
    ll dp[maxn][maxn];
    int a[maxn];
    const int UNSOLVED = 0;
    
    ll solve(int l, int r) {
        if(l == r) {
            return 0;
        } else {
            if(dp[l][r] != inf*inf) {
                return dp[l][r];
            } else {
                for(int m = l; m < r; m++) {
                    dp[l][r] = min(dp[l][r], solve(l, m) + solve(m + 1, r));
                }
                for(int i = l; i <= r; i++) {
                    dp[l][r] += a[i];
                }
                return dp[l][r];
            }
        }
    }
    
    void main() {
        int n;
        cin >> n;
    
        REP(i,0,n){
            REP(j,0,n)dp[i][j] = inf*inf;
        }
        REP(i, 0, n)cin >> a[i];
        cout << solve(0, n - 1);
    }
}


signed main() {
#ifndef ONLINE_JUDGE
#endif
    
    
    int t = 1;
//    in(t);
    for(int i = 1; i <= t; i++) {
//        cout<<"Case #"<<i<<":";
        SOLVE::main();
        
    }








//    clock_t st = clock();
//    while(clock() - st < 3.0 * CLOCKS_PER_SEC){
//
//    }
    
    
    
    
    
    
    return 0;
}