CodeForces D. Concatenated Multiples

http://codeforces.com/contest/1029/problem/D

You are given an array aa, consisting of nn positive integers.

Let's call a concatenation of numbers xx and yy the number that is obtained by writing down numbers xx and yy one right after another without changing the order. For example, a concatenation of numbers 1212 and 34563456 is a number 123456123456.

Count the number of ordered pairs of positions (i,j)(i,j) (i≠ji≠j) in array aa such that the concatenation of aiai and ajaj is divisible by kk.

Input

The first line contains two integers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, 2≤k≤1092≤k≤109).

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109).

Output

Print a single integer — the number of ordered pairs of positions (i,j)(i,j) (i≠ji≠j) in array aa such that the concatenation of aiai and ajaj is divisible by kk.

Examples

input

Copy

6 11
45 1 10 12 11 7

output

Copy

7

input

Copy

4 2
2 78 4 10

output

Copy

12

input

Copy

5 2
3 7 19 3 3

output

Copy

0

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + 10;
int N, K;
int num[maxn];
map<long long, long long> mp[15];

int main() {
    scanf("%d%d", &N, &K);
    for(int i = 1; i <= N; i ++) {
        scanf("%d", &num[i]);
        long long a = num[i];
        for(int j = 1; j <= 10; j ++) {
            a *= 10;
            a %= K;
            mp[j][a] ++;
        }
    }
    long long cnt = 0;
    for(int i = 1; i <= N; i ++) {
        int t = num[i] % K;
        int len = log10(num[i]) + 1;
        cnt += mp[len][(K - t) % K];
        long long x = 1;
        for(int j = 1; j <= len; j ++)
            x = (x * 10) % K;
        if(((num[i] * x) % K + num[i] % K) % K == 0)
            cnt --;
    }
    printf("%I64d\n", cnt);
    return 0;
}