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;
int mod = 1000000007;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
inline long long toll(string s) {
  long long v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
int dcmp(double x, double y) {
  return fabs(x - y) <= 1e-9 ? 0 : x < y ? -1 : 1;
}
void printV(vector<int> v) {
  for (int i = 0; i < v.size(); i++) {
    if (i) {
      cout << " " << v[i];
    } else
      cout << v[i];
  }
}
int fixMod(int a) { return (a % mod + mod) % mod; }
int invalid = 1000 + 3;
int n, m;
int memo[1000 + 5][1000 + 5];
vector<int> v;
bool res = 0;
bool solve(int i, int rem) {
  if (i == n) {
    if (rem == 0) {
      return 1;
    } else {
      return 0;
    }
  }
  if (memo[i][rem] != -1) {
    return memo[i][rem];
  }
  bool take = 0;
  if (rem == invalid) {
    take = solve(i + 1, v[i] % m);
  } else {
    take = solve(i + 1, (v[i] + rem) % m);
  }
  bool leave = solve(i + 1, rem);
  return memo[i][rem] = take | leave;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    v.push_back(t);
  }
  if (n > m) {
    cout << "YES\n";
    return 0;
  }
  memset(memo, -1, sizeof(memo));
  if (solve(0, invalid)) {
    cout << "YES\n";
  } else {
    cout << "NO\n";
  }
  return 0;
}
