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;
using ll = long long;
const int MOD = 1e9 + 7;
const ll INF = 1e18 + 9;
using pi = pair<int, int>;
ll powmod(ll base, ll exp, const ll MOD) {
  ll ans = 1;
  while (exp) {
    if (exp & 1) ans = (ans * base) % MOD;
    base = (base * base) % MOD;
    exp >>= 1;
  }
  return ans;
}
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return (a * b) / gcd(a, b); }
ll modInverse(ll n, const ll p) { return powmod(n, p - 2, p); }
void add(ll &x, ll v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int mxN = 1e6 + 99;
int a[mxN];
int dp[1001][1001];
void test_case() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  bool ok = false;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i] %= m;
    if (a[i] == 0) ok = true;
  }
  if (n > m)
    ok = true;
  else {
    for (int i = 0; i < n; ++i) dp[i + 1][a[i]] = 1;
    for (int i = 1; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        if (dp[i][j]) {
          dp[i + 1][(j + a[i]) % m] = 1;
          dp[i + 1][j] = 1;
        }
      }
    }
    if (dp[n][0]) ok = true;
  }
  if (ok)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
