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;
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<long long int> prime;
void sieve(long long int n) {
  bool bakh[n + 1];
  memset(bakh, true, sizeof(bakh));
  for (long long int p = 2; p * p <= n; p++) {
    if (bakh[p] == true) {
      for (long long int i = p * p; i <= n; i += p) bakh[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (bakh[p]) prime.push_back(p);
}
long long int eulertotient(long long int z) {
  long long int fac = z;
  for (long long int i = 0; prime[i] * prime[i] <= z; i++) {
    if (z % prime[i] == 0) {
      fac -= (fac / prime[i]);
      while (z % prime[i] == 0) z /= prime[i];
    }
  }
  if (z > 1) fac -= (fac / z);
  return fac;
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int lcm(long long int a, long long int b) {
  long long int g = gcd(a, b);
  long long int ans = (a * b) / g;
  return ans;
}
long long int modInverse(long long int a, long long int m) {
  long long int hvf = gcd(a, m);
  if (hvf == 1) return power(a, m - 2, m);
  return -1;
}
void multiply(long long int F[2][2], long long int M[2][2]) {
  long long int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
  long long int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
  long long int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
  long long int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
void powermat(long long int F[2][2], long long int n) {
  if (n == 0 || n == 1) return;
  long long int M[2][2] = {{1, 1}, {1, 0}};
  powermat(F, n / 2);
  multiply(F, F);
  if (n % 2 != 0) multiply(F, M);
}
long long int fib(long long int n) {
  long long int F[2][2] = {{1, 1}, {1, 0}};
  if (n == 0) return 0;
  powermat(F, n - 1);
  return F[0][0];
}
vector<long long int> merge(vector<long long int> v1,
                            vector<long long int> v2) {
  vector<long long int> temp;
  long long int i = 0, j = 0;
  while (i < v1.size() && j < v2.size()) {
    if (v1[i] < v2[j])
      temp.push_back(v1[i]), i++;
    else
      temp.push_back(v2[j]), j++;
  }
  while (i < v1.size()) temp.push_back(v1[i]), i++;
  while (j < v2.size()) temp.push_back(v2[j]), j++;
  return temp;
}
long long int dp[1002][1002], n, m;
long long int ar[1000005];
bool fun(long long int id, long long int val) {
  if (val == 0) return 1;
  if (id == n) return 0;
  if (dp[id][val] != -1) return dp[id][val];
  return dp[id][val] = (fun(id + 1, (val + ar[id]) % m) || fun(id + 1, val));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) cin >> ar[i], ar[i] %= m;
  if (n > m) return cout << "YES", 0;
  memset(dp, -1, sizeof(dp));
  bool ans = fun(0, m);
  if (ans == 1)
    cout << "YES\n";
  else
    cout << "NO\n";
}
