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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110, M = 10, INF = 0x3f3f3f3f;
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int n, m;
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int w[N], s[N];
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int f[N][N][M], g[N][N][M];
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int get_mod(int x) {
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return (x % 10 + 10) % 10;
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}
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int main() {
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cin >> n >> m;
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for (int i = 1; i <= n; i++) {
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cin >> w[i];
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w[i + n] = w[i]; // 破环成链
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}
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for (int i = 1; i <= n * 2; i++) s[i] = s[i - 1] + w[i]; // 预处理前缀和,下标从1开始
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memset(f, 0x3f, sizeof f); // 预求最小，先设最大
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memset(g, -0x3f, sizeof g); // 预求最大，先设最小
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for (int len = 1; len <= n; len++) // 区间DP,先枚举长度
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for (int l = 1; l + len - 1 <= n * 2; l++) { // 枚举起点
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int r = l + len - 1; // 计算终点
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// DP 初始化
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// 边界条件是全部划分为k=1,也就是一块时的结果是多少，根据题意，一整块当然就是区间内容相加后对10取模
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f[l][r][1] = g[l][r][1] = get_mod(s[r] - s[l - 1]);
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// 状态转移
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for (int k = 2; k <= m; k++) // 枚举每个可以划分成的部分
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for (int j = l + k - 2; j <= r - 1; j++) { // 枚举K-1部分时的最后一个终点位置
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f[l][r][k] = min(f[l][r][k], f[l][j][k - 1] * get_mod(s[r] - s[j]));
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g[l][r][k] = max(g[l][r][k], g[l][j][k - 1] * get_mod(s[r] - s[j]));
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}
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}
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// 枚举每个区间长度为n的区间，获取符合长度要求的区间内，划分成m块的最大、最小值
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int mx = -INF, mi = INF;
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for (int i = 1; i <= n; i++) {
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mx = max(mx, g[i][i + n - 1][m]);
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mi = min(mi, f[i][i + n - 1][m]);
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}
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cout << mi << endl;
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cout << mx << endl;
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return 0;
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}
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