You are given a sequence of numbers a1, a2, ..., an, and a number m.

Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.

Input

The first line contains two numbers, n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.

Examples

Input

3 5
1 2 3


Output

YES


Input

1 6
5


Output

NO


Input

4 6
3 1 1 3


Output

YES


Input

6 6
5 5 5 5 5 5


Output

YES

Note

In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.

In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.

In the third sample test you need to choose two numbers 3 on the ends.

In the fourth sample test you can take the whole subsequence.
#include <bits/stdc++.h>
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
string to_str(T str) {
  stringstream stream;
  stream << str;
  return stream.str();
}
template <typename T>
int to_int(T num) {
  int val;
  stringstream stream;
  stream << num;
  stream >> val;
  return val;
}
vector<string> split(const string& s, char delim) {
  vector<string> elems;
  stringstream ss(s);
  string item;
  while (getline(ss, item, delim)) elems.push_back(item);
  return elems;
}
const int dr[] = {-1, -1, 0, 1, 1, 1, 0, -1};
const int dc[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int T2 = 105;
const int T3 = 1005;
const int T4 = 10005;
const int T5 = 100005;
int a[T3];
vector<int> v;
void f(int x) {
  if (!a[x]) a[x] = 1, v.push_back(x);
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, m, x;
  cin >> n >> m;
  for (int i = 0; i < (n); i++) {
    cin >> x;
    x %= m;
    int ln = v.size();
    for (int k = 0; k < ln; k++) f((v[k] + x) % m);
    if (!a[x]) a[x] = 1, v.push_back(x);
    if (a[0]) {
      cout << "YES\n";
      return 0;
    }
  }
  cout << "NO\n";
  return 0;
}
