#include<bits/stdc++.h>
#define endl '\n'
#define x first
#define y second
#define clr(arr,a) memset(arr, a, sizeof(arr))
#define IOS ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<ll, int> Pll;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MOD = 11380;
const int INF = 0x3f3f3f3f;
const int maxn = 1e2+10;
__int128 dp[maxn][maxn];
ll a[maxn]; 
int main(void) {
    int n; cin >> n;
    for (int i = 1; i<=n; ++i) cin >> a[i];
    __int128 t = 1;
    for (int i = 1; i<=n; ++i) 
        for (int j = 1; j<=n; ++j)
            dp[i][j] = 1e30;
    for (int i = 1; i<=n-2; ++i) dp[i][i+2] = t*a[i]*a[i+1]*a[i+2];
    for (int i = 1; i<=n-1; ++i) dp[i][i+1] = 0;
    for (int len = 3; len<n; ++len) 
        for (int l = 1; l+len<=n; ++l) {
            int r = l+len;
            //下面k取l+1和r-1实际上是划分了两个区间，那个长度为2的区间的贡献应该为0，所以上面dp[i][i+1] = 0
            for (int k = l+1; k<r; ++k) dp[l][r] = min(dp[l][r], dp[l][k]+dp[k][r]+t*a[l]*a[r]*a[k]);
        }
    string ans;
    while(dp[1][n]) {
        ans += dp[1][n]%10+'0';
        dp[1][n] /= 10;
    }
    reverse(ans.begin(), ans.end());
    cout << ans << endl;
    return 0;
}