//I use integers, but depending on constraints, it could need long longs
    vector<vector<int>> dp(k, vector<int>(n + 1));
    for (int i = 0; i < k; i++) {
        dp[i][0] = 1;
        
        for (int j = 1; j <= n; j++)
            //For the first block we calculate just prefix products
            if (i == 0) {
                dp[i][j] = dp[i][j - 1] * a[j - 1];
            } else {
                //Start with the last element in the prefix
                int cur = a[j - 1];
                dp[i][j] = max(cur, dp[i - 1][j - 1]);
                
                for (int t = j - 1; t > i; t--) {
                    //Add one more element to the product (moving to beginning of the array)
                    cur *= a[t - 1];
                    //Try to improve current value
                    dp[i][j] = min(dp[i][j], max(cur, dp[i - 1][t - 1]));
                }
            }
    }
    
    int answer = dp[k - 1][n];