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;
template <class T>
void read(T& num) {
  char CH;
  bool F = false;
  for (CH = getchar(); CH < '0' || CH > '9'; F = CH == '-', CH = getchar())
    ;
  for (num = 0; CH >= '0' && CH <= '9';
       num = num * 10 + CH - '0', CH = getchar())
    ;
  F && (num = -num);
}
int stk[70], tp;
template <class T>
inline void print(T p) {
  if (!p) {
    puts("0");
    return;
  }
  while (p) stk[++tp] = p % 10, p /= 10;
  while (tp) putchar(stk[tp--] + '0');
  putchar('\n');
}
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
const int inf = 1e9;
const int N = 1e6 + 20;
const int maxn = 1e3 + 110;
const double eps = 1e-12;
int n, m, a[N], vis[maxn], temp[maxn];
int main() {
  read(n);
  read(m);
  int flag = 0;
  for (int i = 1; i <= n; i++) read(a[i]);
  vis[0] = 1;
  for (int i = 1; i <= n; i++) {
    a[i] %= m;
    int t = (m - a[i]) % m;
    if (vis[t]) {
      flag = 1;
      break;
    }
    memset(temp, 0, sizeof(temp));
    ;
    for (int j = 0; j < m; j++) {
      if (vis[j]) temp[(j + a[i]) % m] = 1;
    }
    for (int j = 0; j < m; j++)
      if (temp[j]) vis[j] = 1;
  }
  if (flag)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
