#include <iostream>
#include <limits.h>
#include <cstring>
using namespace std;
const int N = 110, M = 10;
int a[N], s[N];
int f[N][N][M], g[N][N][M];
int get_mod(int x){
    return (x % 10 + 10) % 10;
}
int main(){
    int n, m;   cin >> n >> m;;
    for(int i = 1; i <= n; i ++ ){
        cin >> a[i];
        a[i + n] = a[i];
    }
    for(int i = 1; i <= 2 * n; i ++ )   s[i] = s[i - 1] + a[i];
    memset(f, 0x3f3f3f, sizeof f);
    memset(g, -0x3f3f3f, sizeof g);
    for(int len = 1; len <= n; len ++ ){
        for(int l = 1; l + len - 1 <= 2 * n; l ++ ){
            int r = l + len - 1;
            for(int k = 1; k <= m; k ++ ){
                if(k == 1)    f[l][r][k] = g[l][r][k] = get_mod(s[r] - s[l - 1]);
                else{
                    for(int j = l + k - 2; j < r; j ++ ){
                        f[l][r][k] = min(f[l][r][k], f[l][j][k - 1] * get_mod(s[r] - s[j]));
                        g[l][r][k] = max(g[l][r][k], g[l][j][k - 1] * get_mod(s[r] - s[j]));
                    }   
                }
            }
        }
    }
    int maxv = INT_MIN, minv = INT_MAX;
    for(int i = 1; i <= n; i ++ ){
        minv = min(minv, f[i][i + n - 1][m]);
        maxv = max(maxv, g[i][i + n - 1][m]);
    }
    cout << minv << endl;
    cout << maxv << endl;
    return 0;
}