#include <bits/stdc++.h>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT_ON(bit, i) (bit & (1LL << i))

typedef long long LL;
template<typename T> using V = std::vector<T>;
template<typename T> using VV = std::vector<std::vector<T>>;
template<typename T> using VVV = std::vector<std::vector<std::vector<T>>>;
template<typename T> using VVVV = std::vector<std::vector<std::vector<std::vector<T>>>>;

template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline void print_vec(const std::vector<T> &v) { for (int i = 0; i < v.size(); ++i) {  if (i != 0) {std::cout << " ";} std::cout << v[i];} std::cout << "\n"; }
template<class T> inline void print_2vec(const std::vector<std::vector<T>> &v) { for (int i = 0; i < v.size(); ++i) { for (int j = 0; j < v[0].size(); ++j) { if (j != 0) { std::cout << " "; } std::cout << std::setw(13) << std::left << v[i][j]; } std::cout << "\n";}}
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }
template<typename T> T &chmin(T &a, const T &b) { return a = std::min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = std::max(a, b); }
template<class T> inline std::vector<T> unique(std::vector<T> v) { sort(v.begin(), v.end()); v.erase(unique(v.begin(), v.end()), v.end()); return v; }
// 初項s, 交差dのn個の数列の和
long long sum_of_arithmetic_progression(long long s, long long d, long long n) { return n * (2 * s + (n - 1) * d) / 2; }

const int INF = 1 << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy4 = {0, -1, 0, 1}, dx4 = {1, 0, -1, 0};
const std::vector<int> dy8 = { 0, -1, 0, 1, 1, -1, -1, 1 }, dx8 = { 1, 0, -1, 0, 1, 1, -1, -1 };

using namespace std;


class CumulativeSum {

public:
    std::vector<long long> memo1;               // 1次元用の累積和表
    std::vector<std::vector<long long>> memo2;  // 2次元用の累積和表

    CumulativeSum() {}

    void build(const std::vector<long long> &line) {
        this->memo1.assign(line.size() + 1, 0);

        for (long long i = 0; i < line.size(); ++i) {
            this->memo1[i + 1] = this->memo1[i] + line[i];
        }
    }

    // 2次元の累積和表を作成
    void build(const std::vector<std::vector<long long>> &board) {
        int height = board.size();
        int width = board[0].size();
        this->memo2.assign(height + 1, std::vector<long long>(width + 1, 0));

        for (int y = 0; y < height; ++y) {
            for (int x = 0; x < width; ++x) {
                memo2[y + 1][x + 1] = board[y][x] + memo2[y + 1][x];
            }
            for (int x = 0; x < width; ++x) {
                memo2[y + 1][x + 1] += memo2[y][x + 1];
            }
        }
    }

    long long sum(int left, int right) {
        return this->memo1[right] - this->memo1[left];
    }

    /*
    (y1, x1)から(y2, x2)の合計を返す．(y2, x2)は含まない
    座標はmemoを作成したboard準拠
    (ex, sum(0, 0, 2, 2)なら(0, 0), (0, 1), (1, 0), (1, 1)の合計を返す)
    */
    long long sum(int y1, int x1, int y2, int x2) {
        return this->memo2[y2][x2] - this->memo2[y2][x1] - this->memo2[y1][x2] + this->memo2[y1][x1];
    }
};

int main() {
    cin.tie(0);
	ios::sync_with_stdio(false);

	int N;
	cin >> N;
	V<LL> A(N);
	FOR(i, 0, N) {
	    cin >> A[i];
	}

	CumulativeSum cs;
	cs.build(A);


	VV<LL> dp(N + 1, V<LL>(N + 1, LONG_LONG_MAX));
	FOR(i, 0, N) {
	    dp[i][i] = 0;
	}
	FOR(i, 0, N - 1) {
        dp[i][i + 1] = A[i] + A[i + 1];
    }

    FOR(len, 3, N + 1) {
	    FOR(left, 0, N) {
	        int right = left + len - 1;
	        if (right >= N) {
                continue;
	        }
	        FOR(m, left, right) {
	            chmin(dp[left][right], dp[left][m] + dp[m + 1][right] + cs.sum(left, right + 1));
	        }
	    }
	}

	print(dp[0][N - 1]);

	return 0;
}

Measured memory_bytes: 1536