#include <bits/stdc++.h>

using namespace std;

const int N = 550;
const int mod1 = 998244353;
const int mod2 = 1e9 + 7;
const int mod3 = 1e9 + 9;

int num1, num2, num3;
int n, k;
int m1, m2, m3;
int a1[N], a2[N], a3[N];
int ans = 0;
char a[100010];

int add(int x, int y, int mod) {
    return x + y >= mod ? x + y - mod : x + y;
}

void solve() {
    for (int i = 1; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {
            for (int k = j; k <= n; ++k) {
                int x1 = add(add(a1[i], a1[j], mod1), a1[k], mod1);
                int x2 = add(add(a2[i], a2[j], mod2), a2[k], mod2);
                int x3 = add(add(a3[i], a3[j], mod3), a3[k], mod3);
                if (x1 == m1 && x2 == m2 && x3 == m3) {
                    ++ans;
                }
            }
        }
    }
}

void divv(int l, int r) {
    if (r - l + 1 == k) return;
    divv(l, r - k);
    int pl = r - 2 * k + 1, pr = r - k;
    // cerr << pl << ' ' << pr << '\n';
    for (int i = pr + 1; i <= r; i++) a[i] = a[i] + a[i - k];
    for (int i = pl; i <= pr; i++) a[i] = 0;
    for (int i = r; i > pr; i--)
        if (a[i] > 9) a[i] -= 10, a[i - 1]++;
    while (a[pr] == 1) {
        a[pr] = 0;
        a[r]++;
        for (int i = r; a[i] > 9; i--) a[i] -= 10, a[i - 1]++;
    }
    return;
}

int main() {
    scanf("%d%d", &n, &k);
    
    for (int i = 1; i <= n; ++i) {
        // char ch
        scanf("%s", a + 1);
        int len = strlen(a + 1), tlen = (len / k + (len % k > 0)) * k;

        for (int i = len; i; i--) a[tlen - (len - i)] = a[i];
        for (int i = 1; i <= tlen - len; i++) a[i] = '0';
        for (int i = 1; i <= tlen; ++i) a[i] -= '0';
        // cerr << '#' << ' ' ;
        // for (int i = 1; i <= tlen; ++i) {
        //     cerr << (int)a[i] << ' ';
        // }
        // cerr << '\n';
        divv(1, tlen);

        for (int j = tlen - k + 1; j <= tlen; ++j) {
            // cerr << (int)a[j] << ' ';
            a1[i] = (1ll * a1[i] * 10 + 1ll * a[j]) % mod1;
            a2[i] = (1ll * a2[i] * 10 + 1ll * a[j]) % mod2;
            a3[i] = (1ll * a3[i] * 10 + 1ll * a[j]) % mod3;
        }
        // cerr << a1[i] <<'\n';
    }
    num1 = num2 = num3 = 0;
    for (int i = 1; i <= k; ++i) {
        num1 = (1ll * num1 * 10 + 1ll * 9) % mod1;
        num2 = (1ll * num2 * 10 + 1ll * 9) % mod2;
        num3 = (1ll * num3 * 10 + 1ll * 9) % mod3;
    }
    solve();
    m1 = add(m1, num1, mod1), m2 = add(m2, num2, mod2), m3 = add(m3, num3, mod3);
    solve();
    m1 = add(m1, num1, mod1), m2 = add(m2, num2, mod2), m3 = add(m3, num3, mod3);
    solve();
    m1 = add(m1, num1, mod1), m2 = add(m2, num2, mod2), m3 = add(m3, num3, mod3);
    solve();
    printf("%d\n", ans);
    return 0;
}