#include <bits/stdc++.h>

using namespace std;
const int N = 55;

typedef __int128 LL;
const LL INF = 1e38;

LL read() {
    LL x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

void write(LL x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

LL f[N][N];
LL w[N];
int n;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) w[i] = read();

    // 初始化
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            // 两个点之间最小差距是2，比如[1,3]这里面就有三角形，需要求解，如果小于2，比如[1,2]
            // 无法构成三角形，就不需要求解，无法构成的情况贡献值是0，也是递推的起点，否则，全都是INF
            // 就无法推起来了
            j - i < 2 ? f[i][j] = 0 : f[i][j] = INF;

    for (int len = 3; len <= n; len++) { // 区间范围最少是3个，否则无法构建三角形
        for (int l = 1; l + len - 1 <= n; l++) {
            int r = l + len - 1;
            for (int k = l + 1; k < r; k++) // l<k<r才能构建三角形
                f[l][r] = min(f[l][r], f[l][k] + f[k][r] + w[l] * w[k] * w[r]);
        }
    }
    // 输出
    write(f[1][n]);
    return 0;
}