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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)               //repマクロ
#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i) //多倍長
#define all(x) (x).begin(), (x).end()                              //先頭と最後を渡す
#define SZ(x) ((int)(x).size())                                    //sizeの返り値をunsigned int型からint型にする
//短くするだけ
#define pb push_back
#define mp make_pair
// 名前の変更
#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define next asdnext
#define prev asdprev

#define cauto const auto &                                      //これいる？
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end()); //なにこれ
#define itn int                                                 //これいる？
// longlong型を省略して
typedef long long lint;

/* 最大値を更新するテンプレート */
template <class T>
bool umax(T &a, const T &b)
{
  if (a < b)
  {
    a = b;
    return 1;
  }
  return 0;
}
/* 最小値を更新するテンプレ */
template <class T>
bool umin(T &a, const T &b)
{
  if (b < a)
  {
    a = b;
    return 1;
  }
  return 0;
}

int main()
{
  int a[3];
  scanf("%d %d %d", a[0], a[1], a[2]);
  if (a[0] > a[1])
  {
    int temp = a[0];
    a[0] = a[1];
    a[1] = temp;
  }
  if (a[0] > a[2])
  {
    int temp = a[0];
    a[0] = a[2];
    a[2] = temp;
  }
  if (a[0] == 5 && a[1] == 7 && a[2] == 7)
  {
    printf("YES\n");
  }
  else
  {
    printf("NO\n");
  }
}