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 <bits/stdc++.h>
#define range(i, l, r) for(int (i) = (l); (i) < (r); (i)++)
#define reversed_range(i, l, r) for (int (i) = (r) - 1; (i) >= l; (i)--)
using namespace std;
template <typename T>
using vec = vector<T>;
using lint = long long;
using ulint = unsigned long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;

template <typename S, typename T>
ostream& operator <<(ostream& os, pair<S, T> p) {
    os << "(";
    os << p.first << ", " << p.second;
    return os << ")";
}

template <typename T>
ostream& operator <<(ostream& os, vec<T> v) {
    os << "[";
    for (int i = 0; i < v.size() - 1; i++) {
        os << v.at(i) << ", ";
    }
    return os << v.at(v.size() - 1) << "]";
}

template <typename T>
ostream& operator <<(ostream& os, set<T>& s) {
    os << "{";
    if (s.begin() == s.end()) return os << "}";
    auto it_first_item = s.begin();
    cout << *it_first_item;
    for (auto it = ++it_first_item; it != s.end(); it++) {
        cout << ", " << *it;
    }
    return cout << "}";
}

template <typename T>
ostream& operator <<(ostream& os, unordered_set<T>& s) {
    os << "{";
    if (s.begin() == s.end()) return os << "}";
    auto it_first_item = s.begin();
    cout << *it_first_item;
    for (auto it = ++it_first_item; it != s.end(); it++) {
        cout << ", " << *it;
    }
    return cout << "}";
}

template <typename K, typename V>
ostream& operator <<(ostream& os, map<K, V> m) {
    os << "{";
    if (m.begin() == m.end()) return os << "}";
    auto it_first_item = m.begin();
    cout << it_first_item->first << ": " << it_first_item->second;
    for (auto it = ++it_first_item; it != m.end(); it++) {
        cout << ", " << it->first << ": " << it->second;
    }
    return os << "}";
}

template <typename K, typename V>
ostream& operator <<(ostream& os, unordered_map<K, V> m) {
    os << "{";
    if (m.begin() == m.end()) return os << "}";
    auto it_first_item = m.begin();
    cout << it_first_item->first << ": " << it_first_item->second;
    for (auto it = ++it_first_item; it != m.end(); it++) {
        cout << ", " << it->first << ": " << it->second;
    }
    return os << "}";
}

lint pow(lint num, lint e, lint MOD) {
    lint res = 1;
    lint cur_num = num;
    while (e) {
        if (e & 1) {
            res *= cur_num;
            res %= MOD;
        }
        cur_num *= cur_num;
        cur_num %= MOD;
        e >>= 1;
    }
    return res;
}

int main() {
    cin.tie(0); cout.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    lint A, B, C;
    cin >> A >> B >> C;
    map<lint, lint> count;
    count[A] += 1;
    count[B] += 1;
    count[C] += 1;
    cout << (count[5] == 2 && count[7] == 1 ? "YES" : "NO") << "\n";
}