Codeforce Round #213 Div2 A

A. Good Number

time limit per test 1 second

memory limit per test 256 megabytes

 

Let's call a number k-good if it contains all digits not exceeding k (0, ..., k). You've got a number k and an array a containing n numbers. Find out how many k-good numbers are in a (count each number every time it occurs in array a).

Input

The first line contains integers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 9). The i-th of the following n lines contains integer a_i without leading zeroes (1 ≤ a_i ≤ 10⁹).

Output

Print a single integer — the number of k-good numbers in a.

Sample test(s)

Input

10 6 
1234560 
1234560 
1234560 
1234560 
1234560 
1234560 
1234560 
1234560 
1234560 
1234560

Output

10

Input

2 1 
1 
10

Output

1

#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 10001
#define INF 1000000000
#define ll long long
char s[maxn];
int a[maxn];
int vis[maxn];
ll gcd(ll n, ll m){ return m ? gcd(m, n%m) : n; }
ll n, m, b, c, d, p,t1, t2, cnt1, cnt2, add, sub,ans;
int main(){
    int t, cas = 1;
    scanf("%d%d", &n,&m);
    ans = 0;
    while (n--){
        scanf("%s", s);
        memset(vis, 0, sizeof vis);
        int i;
        for (i = 0; i < strlen(s)-1; i++)if (s[i] != '0')break;
        for (i = 0; i < strlen(s); i++){
            vis[s[i] - '0'] = 1;
        }
        t = 0;
        for (int i = 0; i <=m; i++)if (vis[i])t++;
        if (t == m+1)ans++;
    }
    printf("%d\n", ans);
    return 0;
}