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#include <bits/stdc++.h>
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using namespace std;
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const int N = 30;
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const int INF = 0x3f3f3f3f;
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int a[N];
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bool st[N];
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int n, k;
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int cnt;
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vector<int> path;
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bool isPrime(int n) {
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if (n < 2) return false;
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for (int i = 2; i <= n / i; i++)
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if (n % i == 0) return false;
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return true;
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}
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/**
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* 功能:遍历每个箱子,向箱子里装卡片
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* @param step 走在第几个箱子面前
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* @param sum 已经获得的数字和
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* @param start 从哪个数字可以开始进行选择
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* 举个栗子：
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* 3+7+19
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* 3+19+7
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* 7+3+19
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* 7+19+3
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* 19+3+7
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* 19+7+3 这样就出来6个，肯定不行，需要考虑只出1个！
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* 方法就是“从上一次选择的数字后面一位进行再次选择！”
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* @return 获取可行方法数
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*/
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void dfs(int step, int sum, int start) {
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//当走到第k+1个虚拟箱子面前，表示前面k个箱子已经装完
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if (step == k + 1) {
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if (isPrime(sum)) {
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cnt++;
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for (int i = 0; i < path.size(); i++)
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cout << path[i] << " ";
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cout << endl;
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}
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return;
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}
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//在没有走完箱子时，考虑当前箱子放哪张卡片
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for (int i = start; i <= n; i++) {
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if (!st[i]) {
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st[i] = true;
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path.push_back(i);
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dfs(step + 1, sum + a[i], i + 1);
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path.pop_back();
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st[i] = false;
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}
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}
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}
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/*
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4 3
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3 7 12 19
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答案：1
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*/
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int main() {
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cin >> n >> k;
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for (int i = 1; i <= n; i++) cin >> a[i];
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dfs(1, 0, 1);
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cout << cnt << endl;
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return 0;
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}
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