You are given an array a, consisting of n positive integers.

Let's call a concatenation of numbers x and y the number that is obtained by writing down numbers x and y one right after another without changing the order. For example, a concatenation of numbers 12 and 3456 is a number 123456.

Count the number of ordered pairs of positions (i, j) (i ≠ j) in array a such that the concatenation of a_i and a_j is divisible by k.

Input

The first line contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5, 2 ≤ k ≤ 10^9).

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

Output

Print a single integer — the number of ordered pairs of positions (i, j) (i ≠ j) in array a such that the concatenation of a_i and a_j is divisible by k.

Examples

Input

6 11
45 1 10 12 11 7


Output

7


Input

4 2
2 78 4 10


Output

12


Input

5 2
3 7 19 3 3


Output

0

Note

In the first example pairs (1, 2), (1, 3), (2, 3), (3, 1), (3, 4), (4, 2), (4, 3) suffice. They produce numbers 451, 4510, 110, 1045, 1012, 121, 1210, respectively, each of them is divisible by 11.

In the second example all n(n - 1) pairs suffice.

In the third example no pair is sufficient.
#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  bool first = true;
  os << "{";
  for (unsigned int i = 0; i < v.size(); i++) {
    if (!first) os << ",";
    os << v[i];
    first = false;
  }
  return os << "}";
}
template <typename T>
inline ostream& operator<<(ostream& os, const set<T>& v) {
  bool first = true;
  os << "{";
  for (typename set<T>::const_iterator ii = v.begin(); ii != v.end(); ++ii) {
    if (!first) os << ",";
    os << *ii;
    first = false;
  }
  return os << "}";
}
template <typename T>
inline ostream& operator<<(ostream& os, const multiset<T>& v) {
  bool first = true;
  os << "{";
  for (typename multiset<T>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ",";
    os << *ii;
    first = false;
  }
  return os << "}";
}
template <typename T1, typename T2>
inline ostream& operator<<(ostream& os, const map<T1, T2>& v) {
  bool first = true;
  os << "{";
  for (typename map<T1, T2>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ",";
    os << *ii;
    first = false;
  }
  return os << "}";
}
template <typename T>
bool isPrime(T n) {
  T i, lmt = sqrt(n);
  for (i = 2; i <= lmt; i++)
    if (n % i == 0) return false;
  return true;
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
T BigMod(T b, T p, T m) {
  if (p == 0)
    return 1 % m;
  else if (p & 1)
    return ((b % m) * (BigMod(b, p - 1, m) % m)) % m;
  else {
    T s = BigMod(b, p / 2, m);
    return ((s % m) * (s % m)) % m;
  }
}
template <typename T>
T power(T b, T p) {
  if (p == 0)
    return 1;
  else if (p & 1)
    return b * power(b, p - 1);
  else {
    T ret = power(b, p / 2);
    return ret * ret;
  }
}
template <typename T>
T ModInv(T b, T m) {
  if (isPrime(b))
    return BigMod(b, m - 2, m) % m;
  else if (gcd(b, m) == 1)
    return BigMod(b, m - 1, m) % m;
}
template <typename T>
T egcd(T a, T b, T& x, T& y) {
  T d;
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  } else {
    T x1, y1;
    d = egcd(b % a, a, x1, y1);
    x = y1 - (b / a) * x1;
    y = x1;
  }
  return d;
}
int toint(string s) {
  int i, j, l;
  int num;
  l = s.size();
  for (i = l - 1, j = 0, num = 0; i >= 0; i--, j++) {
    num += (s[i] - 48) * power(10, j);
  }
  return num;
}
const double pi = acos(-1);
const int inf = (1 << 30);
const long long INF = (1LL << 62) - 1;
const int mod = (int)1e9 + 7;
const long long MOD = (long long)1e9 + 7;
const double EPS = 1e-9;
const int dx8[] = {-1, 0, +1, -1, +1, -1, 0, +1};
const int dy8[] = {+1, +1, +1, 0, 0, -1, -1, -1};
const int dx4[] = {0, -1, +1, 0};
const int dy4[] = {+1, 0, 0, -1};
const int dx2[] = {+1, 0};
const int dy2[] = {0, +1};
const int dxkn[] = {1, -1, 1, -1, 2, -2, -2, 2};
const int dykn[] = {2, 2, -2, -2, 1, 1, -1, -1};
vector<long long> val[12];
long long a[200009];
long long md[200009];
long long p[11] = {1,           10LL,         100LL,        1000LL,
                   10000LL,     100000LL,     1000000LL,    10000000LL,
                   100000000LL, 1000000000LL, 10000000000LL};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  unsigned long long k, ans, num, len, m;
  int n, i, j, l, r;
  cin >> n >> k;
  for (i = 1; i < 11; i++) {
    p[i] %= k;
  }
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    md[i] = a[i] % k;
    for (j = 1; j <= 10; j++) {
      val[j].push_back((md[i] * p[j]) % k);
    }
  }
  for (i = 1; i < 11; i++) {
    sort(val[i].begin(), val[i].end());
  }
  for (i = 1, ans = 0LL; i <= n; i++) {
    num = a[i];
    len = 0;
    while (num) {
      len++;
      num /= 10LL;
    }
    m = md[i];
    if (md[i]) m = k - m;
    l = lower_bound(val[len].begin(), val[len].end(), m) - val[len].begin();
    r = upper_bound(val[len].begin(), val[len].end(), m) - val[len].begin();
    ans += (r - l);
    if ((md[i] * p[len] + md[i]) % k == 0) {
      ans--;
    }
  }
  cout << ans << "\n";
  return 0;
}
