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)

#include <bits/stdc++.h>
#include <stdio.h>
#include <iostream>
#include <cstdio>
#include <bitset>
#include <algorithm>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <string>
#include <string.h>
#include <cmath>
#include <utility>
#include <functional>
#include <map>
#include <set>
#include <cctype>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <cstring>
using namespace std;
using ll = long long;
#define ALL(a) (a).begin(), (a).end()

#define FOR(i, a, b) for (long long int i = (a); i <= (b); i++)
#define RFOR(i, a, b) for (long long int i = (a); i >= (b); i--)
#define MOD 1000000007
#define LLONG_MAXs 9223372036854775800 / 2

#define ALL(a) (a).begin(), (a).end()

#include <iostream>
#include <cmath>
using namespace std;

bool isPrimeNum(ll x)
{ // 素数である場合 true を返す
	if (x <= 1)
	{ // 1以下である場合は素数でないことがすぐにわかる
		return false;
	}
	// sqrt( double型 ) は引数の平方根を double型で返すので、int型でキャスト
	int n = (int)sqrt((double)x);
	for (int i = 2; i <= n; i++)
	{
		if (x % i == 0)
		{ // 割り切る整数がある場合、即判定終了
			return false;
		}
	}
	return true; // 割り切る整数がない場合、素数である
}

ll myPow(ll x, ll n, ll m)
{
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return myPow(x * x % m, n / 2, m);
	else
		return x * myPow(x, n - 1, m) % m;
}

constexpr ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
constexpr ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
constexpr ll abs(ll a, ll b)
{
	if (a >= b)
		return a - b;
	if (a < b)
		return b - a;
}
constexpr double dabs(double a, double b)
{
	if (a >= b)
		return a - b;
	if (a < b)
		return b - a;
}
constexpr ll min(ll a, ll b)
{
	if (a >= b)
		return b;
	if (a < b)
		return a;
}

constexpr ll max(ll a, ll b)
{
	if (a >= b)
		return a;
	if (a < b)
		return b;
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

int dx8[8] = {1, 0, -1, 0, 1, -1, 1, -1};
int dy8[8] = {0, 1, 0, -1, 1, 1, -1, -1};

class UnionFind
{
public:
	//親の番号を格納する。親だった場合は-(その集合のサイズ)
	vector<int> Parent;

	//作るときはParentの値を全て-1にする
	//こうすると全てバラバラになる
	UnionFind(int N)
	{
		Parent = vector<int>(N, -1);
	}

	//Aがどのグループに属しているか調べる
	int root(int A)
	{
		if (Parent[A] < 0)
			return A;
		return Parent[A] = root(Parent[A]);
	}

	//自分のいるグループの頂点数を調べる
	int size(int A)
	{
		return -Parent[root(A)]; //親をとってきたい]
	}
	bool issame(int x, int y)
	{
		return root(x) == root(y);
	}
	//AとBをくっ付ける
	bool connect(int A, int B)
	{
		//AとBを直接つなぐのではなく、root(A)にroot(B)をくっつける
		A = root(A);
		B = root(B);
		if (A == B)
		{
			//すでにくっついてるからくっ付けない
			return false;
		}

		//大きい方(A)に小さいほう(B)をくっ付けたい
		//大小が逆だったらひっくり返しちゃう。
		if (size(A) < size(B))
			swap(A, B);

		//Aのサイズを更新する
		Parent[A] += Parent[B];
		//Bの親をAに変更する
		Parent[B] = A;

		return true;
	}
};

long long fac[510000], finv[510000], inv[510000];

// テーブルを作る前処理
void COMinit()
{
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < 510000; i++)
	{
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k)
{
	if (n < k)
		return 0;
	if (n < 0 || k < 0)
		return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

long long modinv(long long a, long long m)
{
	long long b = m, u = 1, v = 0;
	while (b)
	{
		long long t = a / b;
		a -= t * b;
		swap(a, b);
		u -= t * v;
		swap(u, v);
	}
	u %= m;
	if (u < 0)
		u += m;
	return u;
}

void yn(bool flag)
{
	if (flag)
	{
		cout << "Yes" << endl;
	}
	else
	{
		cout << "No" << endl;
	}
	return;
}
void YN(bool flag)
{
	if (flag)
	{
		cout << "YES" << endl;
	}
	else
	{
		cout << "NO" << endl;
	}
	return;
}

std::vector<ll> enum_div(ll n) //nの約数を列挙
{
	std::vector<ll> ret;
	for (ll i = 1; i * i <= n; ++i)
	{
		if (n % i == 0)
		{
			ret.push_back(i);
			if (i != 1 && i * i != n)
			{
				ret.push_back(n / i);
			}
		}
	}
	ret.push_back(n);
	return ret;
}

// グラフセット
struct Edge
{
	ll to;	 // 辺の行き先
	ll weight; // 辺の重み
	Edge(int t, int w) : to(t), weight(w) {}
};
using Graph = vector<vector<Edge>>;

//std::setprecision(30)<<ans<<endl;
int main()
{
	ll N, M;
	cin >> N;
	ll p[N];
	ll rui[N + 1] = {};
	ll dp[N + 1][N + 2] = {};
	FOR(i, 0, N - 1)
	{
		cin >> p[i];
		//dp[i][i + 1] = p[i];
		rui[i + 1] = rui[i] + p[i];
	}

	cout << endl;

	FOR(i, 2, N)
	{
		FOR(j, 0, N)
		{
			if (i + j <= N)
			{
				ll mincost = LLONG_MAXs;
				ll sum = 0;
				FOR(k, 1, i - 1)
				{
					
					mincost = min(mincost, dp[j][k + j] + dp[k + j][j + i]);
					sum += p[j + k - 1];
				}

				dp[j][i + j] = mincost + rui[i + j] - rui[j];
			}
		}
	}



	cout << dp[0][N] << endl;

	return 0;
}