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 ll = long long;
using namespace std;
const long double PI = acos(-1);
const int M = 1000000007;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int mod(int n, int m = M) {
  int ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
int exp(int n, int k, int m = M) {
  if (k == 0) return 1;
  if (k == 1) return n;
  int ax = exp(n, k / 2, m);
  ax = mod(ax * ax, m);
  if (k % 2) ax = mod(ax * n, m);
  return ax;
}
int gcd(int a, int b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
void solution(int testcase) {
  int n, m;
  cin >> n >> m;
  vector<int> cnt(m, 0);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    cnt[x % m]++;
  }
  set<int> dp;
  for (int i = 0; i < m; i++) {
    dp.insert(i);
  }
  for (int i = 0; i < m; i++) {
    while (cnt[i]--) {
      if (dp.size() == 0) break;
      vector<int> rem;
      for (int x : dp) {
        int prv = mod(x - i, m);
        if (dp.count(prv)) continue;
        rem.push_back(x);
      }
      rem.push_back(i);
      int iter = 0;
      for (int x : rem)
        if (dp.count(x)) dp.erase(x), iter++;
      if (iter == 0) break;
    }
  }
  if (!dp.count(0))
    cout << "YES" << '\n';
  else
    cout << "NO" << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int _testcase = 1; _testcase <= t; _testcase++) solution(_testcase);
}
