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;
void fun() {}
int mod = 998244353;
long long powe(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
int ad(int x, int y) { return (x % mod + y % mod) % mod; }
int sub(int x, int y) { return (x % mod - y % mod + mod) % mod; }
int mul(int x, int y) { return ((x % mod) * (y % mod)) % mod; }
int divi(int x, int y) { return (mul(x, powe(y, mod - 2))); }
const int N = 1000005;
int n, m;
int dp[1001][1001][2];
int a[N];
int hsh[N];
int solve(int ind, int sum, int f) {
  if (sum == 0 && f == 1) return 1;
  if (ind == m) {
    return (sum == 0 && f == 1);
  }
  if (dp[ind][sum][f] != -1) return dp[ind][sum][f];
  int ans = 0;
  int x = sum;
  for (int i = 0; i <= hsh[ind]; i++) {
    ans |= solve(ind + 1, x % m, (i >= 1));
    if (ans == 1) return dp[ind][sum][f] = ans;
    x += ind;
    x %= mod;
  }
  return dp[ind][sum][f] = ans;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  fun();
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] %= m;
    hsh[a[i]]++;
  }
  memset(dp, -1, sizeof(dp));
  if (hsh[0] > 0) {
    cout << "YES"
         << "\n";
    return 0;
  }
  if (solve(1, 0, 0))
    cout << "YES"
         << "\n";
  else
    cout << "NO"
         << "\n";
}
