Dawid has four bags of candies. The i-th of them contains a_i candies. Also, Dawid has two friends. He wants to give each bag to one of his two friends. Is it possible to distribute the bags in such a way that each friend receives the same amount of candies in total?

Note, that you can't keep bags for yourself or throw them away, each bag should be given to one of the friends.

Input

The only line contains four integers a_1, a_2, a_3 and a_4 (1 ≤ a_i ≤ 100) — the numbers of candies in each bag.

Output

Output YES if it's possible to give the bags to Dawid's friends so that both friends receive the same amount of candies, or NO otherwise. Each character can be printed in any case (either uppercase or lowercase).

Examples

Input


1 7 11 5


Output


YES


Input


7 3 2 5


Output


NO

Note

In the first sample test, Dawid can give the first and the third bag to the first friend, and the second and the fourth bag to the second friend. This way, each friend will receive 12 candies.

In the second sample test, it's impossible to distribute the bags.
#include <bits/stdc++.h>
using namespace std;
clock_t tp = clock();
long long int power1(long long int x, long long int y) {
  long long int res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  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 ncr(long long int n, long long int r) {
  if (n < r) return 0;
  long long int p = 1, k = 1;
  if (n - r < r) r = n - r;
  if (r != 0) {
    while (r) {
      p *= n;
      k *= r;
      long long int g = gcd(p, k);
      p /= g;
      k /= g;
      n--;
      r--;
    }
  } else
    p = 1;
  return p;
}
long long int lcm(long long int a, long long int b) {
  return (a / gcd(a, b) * b);
}
bool compare_incresing(pair<long long int, long long int> l1,
                       pair<long long int, long long int> l2) {
  if (l1.first != l2.first)
    return (l1.first < l2.first);
  else
    return (l1.second < l2.second);
}
long long int n, e[100005][2], par[15], a[100005];
vector<long long int> ed[100005];
long long int parent(long long int n) {
  if (par[n] > 0) return par[n];
  long long int x = 0;
  for (auto it : ed[n]) {
    x = it;
    if (parent(x) != -1) break;
  }
  return par[n] = x;
}
long long int find(long long int n, long long int an) {
  long long int cnt = 0;
  for (auto it : ed[n]) {
    if (it == parent(n)) continue;
    cnt += find(it, an);
  }
  if (an > a[n - 1]) cnt += a[n - 1];
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tt = 1;
  while (tt--) {
    int a[4];
    for (long long int i = 0; i < 4; i++) cin >> a[i];
    for (int i = 0; i < 16; i++) {
      int x = i, sum1 = 0, sum2 = 0, ii = 0;
      while (x) {
        if (x % 2)
          sum1 += a[ii++];
        else
          sum2 += a[ii++];
        x /= 2;
      }
      while (ii < 4) sum2 += a[ii++];
      if (sum1 == sum2) {
        cout << "YES";
        return 0;
      }
    }
    cout << "NO";
  }
  return 0;
}
