Problem:
Score : 100 points

Problem Statement

Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.

To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Constraints

1≦A,B,C≦10

Input

The input is given from Standard Input in the following format:

A B C

Output

If it is possible to construct a Haiku by using each of the phrases once, print YES (case-sensitive). Otherwise, print NO.

Sample Input 1

5 5 7

Sample Output 1

YES

Using three phrases of length 5, 5 and 7, it is possible to construct a Haiku.

Sample Input 2

7 7 5

Sample Output 2

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstring>
#include <map>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <list>
#include <numeric>
#include <stdio.h>
#include <string>
#include <cstdlib>
#include <math.h>
#include <stack>
#include <climits>
#include <bitset>
#include <utility>


using namespace std;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
// const ll INF = 1LL <<60;
const int INF = 10000;

//最小公倍数
ll gcd(ll x, ll y) {
  ll tmp = 0;
  if (x < y){
    tmp=x;
    x=y;
    y=tmp;
  }
  while (y > 0) {
    ll r = x % y;
    x = y;
    y = r;
  }
  return x;
}

//最大公倍数
ll lcm(ll x,ll y){
  return x/gcd(x,y)*y;
}

//階乗
ll kaijo(ll k){
    ll sum = 1;
    for (ll i = 1; i <= k; ++i)
    {
        sum *= i;
        sum%=1000000000+7;
    }
    return sum;
}

//for(int i = ; i < ; i++){}

ll lmax(ll s,ll t){
  if(s>t){
    return s;
  }
  else{
    return t;
  }
}

ll lmin(ll s,ll t){
  if(s<t){
    return s;
  }
  else{
    return t;
  }
}
// ここから開始

int main(){
  int a[3];
  cin>>a[0]>>a[1]>>a[2];
  sort(a,a+3);
  if(a[0]==5&&a[1]==5&&a[2]==7) cout<<"YES"<<endl;
  else cout<<"NO"<<endl;
  return 0;
}