Codeforces Round 213 (Div. 2)
翻译中…
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 $*a*$).
翻译中…
Input
翻译中…
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 ($1 ≤ *a*_{*i*} ≤ 10^{9}$).
翻译中…
Output
翻译中…
Print a single integer — the number of $*k*$-good numbers in $*a*$.
翻译中…
样例
Input #1
10 6
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
Output #1
10
InputCopy
OutputCopy
Input #2
2 1
1
10
Output #2
1
InputCopy
OutputCopy
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