// Problem: No.2090 否定論理積と充足可能性
// Contest: yukicoder
// URL: https://yukicoder.me/problems/no/2090
// Memory Limit: 512 MB
// Time Limit: 2000 ms

#include <bits/stdc++.h>

#define fastio ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define dbg(x) cout << #x << " = " << (x) << "\n";
#define popcount(x) __builtin_popcountll((x))
#define all(v) (v).begin(), (v).end()
#define pb emplace_back
#define x first
#define y second

using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;

const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

bool check(string str) { return !(str.length() == 1 && str[0] == '0'); };

bool nand(bool x, bool y) { return !x || !y; }

void solve() {
  int n = 6;
  vector<string> s(n);
  for (int i = 0; i < n; i++) cin >> s[i];
  vector<bool> t(n);
  for (int i = 0; i < n; i++) t[i] = check(s[i]);
  bool f = nand(nand(nand(t[0], t[1]), t[2]), nand(nand(t[3], t[4]), t[5]));
  if (f)
    cout << "YES\n";
  else
    cout << "NO\n";
}

int main() {
  fastio;
  int t = 1;
  // cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}