Problem: Mishka is decorating the Christmas tree. He has got three garlands, and all of them will be put on the tree. After that Mishka will switch these garlands on.

When a garland is switched on, it periodically changes its state — sometimes it is lit, sometimes not. Formally, if i-th garland is switched on during x-th second, then it is lit only during seconds x, x + ki, x + 2ki, x + 3ki and so on.

Mishka wants to switch on the garlands in such a way that during each second after switching the garlands on there would be at least one lit garland. Formally, Mishka wants to choose three integers x1, x2 and x3 (not necessarily distinct) so that he will switch on the first garland during x1-th second, the second one — during x2-th second, and the third one — during x3-th second, respectively, and during each second starting from max(x1, x2, x3) at least one garland will be lit.

Help Mishka by telling him if it is possible to do this!

Input

The first line contains three integers k1, k2 and k3 (1 ≤ ki ≤ 1500) — time intervals of the garlands.

Output

If Mishka can choose moments of time to switch on the garlands in such a way that each second after switching the garlands on at least one garland will be lit, print YES.

Otherwise, print NO.

Examples

Input

2 2 3


Output

YES


Input

4 2 3


Output

NO

Note

In the first example Mishka can choose x1 = 1, x2 = 2, x3 = 1. The first garland will be lit during seconds 1, 3, 5, 7, ..., the second — 2, 4, 6, 8, ..., which already cover all the seconds after the 2-nd one. It doesn't even matter what x3 is chosen. Our choice will lead third to be lit during seconds 1, 4, 7, 10, ..., though.

In the second example there is no way to choose such moments of time, there always be some seconds when no garland is lit.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
long long binary_expo(long long base, long long expo) {
  long long res = 1;
  while (expo) {
    if (expo % 2) {
      res = mul(res, base);
    }
    expo /= 2;
    base = mul(base, base);
  }
  return res;
}
void solve() {
  long long k1, k2, k3;
  cin >> k1 >> k2 >> k3;
  long long maxi = max({k1, k2, k3});
  if (k1 == 1 || k2 == 1 || k3 == 1) {
    cout << "YES" << '\n';
    return;
  }
  if ((k1 == 2 and k2 == 2) or (k3 == 2 and k1 == 2) or (k2 == 2 and k3 == 2)) {
    cout << "YES" << '\n';
    return;
  }
  if (k1 == 3 and k2 == 3 and k3 == 3) {
    cout << "YES" << '\n';
    return;
  }
  if ((k1 == 2 and k2 == 4 and k3 == 4) or (k1 == 4 and k2 == 2 and k3 == 4) or
      (k1 == 4 and k2 == 4 and k3 == 2)) {
    cout << "YES" << '\n';
    return;
  } else {
    cout << "NO" << '\n';
  }
}
signed main() {
  long long t;
  ios_base::sync_with_stdio(false);
  t = 1;
  while (t--) {
    solve();
  }
}