// Problem: #195. 【FTOJ Round #29】石蒜（manzanita）
// Contest: 
// URL: http://192.168.30.9/problem/195
// Memory Limit: 1024 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define rep1(i, l, r) for (int i = l; i <= int(r); ++i)
#define rep2(i, l, r) for (int i = l; i >= int(r); --i)
#define rer(i, l, r, a) rep1(i, l, r) read(a[i])
#define ptc putchar
#define il inline
#define eb emplace_back
#define ef emplace_front
#define mp make_pair
#define fst first
#define snd second
#define sqr(x) ((x) * (x))
#define ls(x) x << 1
#define rs(x) x << 1 | 1
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 bll;
// typedef unsigned __int128 ubll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int MAXN = 2e4 + 10, LOGN = log2(MAXN) + 1, inf = ~0U >> 2, INF = ~0U >> 1;
const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
namespace stupid_lrc {
	template <typename T> il bool read(T &x) {
		x = 0; bool f = false; char ch;
 		while (!isdigit(ch = getchar())) {
			f ^= !(ch ^ '-');
			if (ch == EOF) return false;
		}
		while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch & 15), ch = getchar();
		if (f) x = ~x + 1; return true;
	}
	il int read() {int x; read(x); return x;}
	
	template <typename T> il bool read(pair <T, T> &x) {return read(x.fst) && read(x.snd);}

	template <typename T> il void gmin(T &x, T y) {x = x < y ? x : y;}

 	template <typename T> il void gmax(T &x, T y) {x = x > y ? x : y;}

	template <typename T, typename ...L>
	il bool read(T &x, L &...y) {return read(x) && read(y...);}

	template <typename T> il T lowbit(const T &x) {return x & -x;}
}
using namespace stupid_lrc;
int n, k, cnt, v[MAXN], ans[510]; char s[MAXN];
const int mod = 1145140019;

int main() {
	read(n, k);
	rep1(i, 1, n) {
		memset(v, 0, sizeof v); scanf("%s", s); int p = strlen(s);
		rep1(i, 0, p - 1) v[i] = s[p - i - 1] - '0';
		int lst = p - 1;
		while (lst >= k) {
			rep1(j, k, lst) v[j % k] += v[j], v[j] = 0;
			rep1(j, 0, lst) v[j + 1] += v[j] / 10, v[j] %= 10;
			rep2(j, lst, 0) if (v[j]) {
				lst = j;
				break;
			}
		}
		rep2(j, k - 1, 0) ans[i] = (ans[i] * 10ll + v[j]) % mod;
	} int M = 0;
	rep1(i, 1, k) M = (M * 10ll + 9) % mod;
	rep1(i, 1, n) rep1(j, i, n) rep1(k, j, n) {
		ll p = ans[i]; p += ans[j]; p += ans[k]; p %= mod;
		if (p == 0 || p == M || p == M * 2ll % mod || p == M * 3ll % mod) ++cnt;
	} printf("%d", cnt);
	return 0;
}