Problem:
Score : 100 points

Problem Statement

You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".

Constraints

0 \leq N_1, N_2, N_3, N_4 \leq 9

N_1, N_2, N_3 and N_4 are integers.

Input

Input is given from Standard Input in the following format:

N_1 N_2 N_3 N_4

Output

If N_1, N_2, N_3 and N_4 can be arranged into the sequence of digits "1974", print YES; if they cannot, print NO.

Sample Input 1

1 7 9 4

Sample Output 1

YES

We can get 1974 by swapping N_2 and N_3.

Sample Input 2

1 9 7 4

Sample Output 2

YES

We already have 1974 before doing anything.

Sample Input 3

1 2 9 1

Sample Output 3

NO

Sample Input 4

4 9 0 8

Sample Output 4

NO

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

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <iterator>  // std::back_inserter()
#include <set>
#include <map>
#include <algorithm>  // std::copy()
#include <functional>  // std::greater<T>()
#include <utility>  // std::swap()
#include <numeric>  // accumulate(ALL(vec), 0)  0 は初期値
#include <bitset>  // static_cast<std::bitset<8>>
#include <sstream>  // std::stringstream
#include <cmath>
#include <climits>  // INT_MIN
#include <cctype>  // std::isdigit()
#include <iomanip>  // std::setprecision()
using namespace std;

#define ALL(obj) (obj).begin(), (obj).end()
#define REP(i, n) for (int i=0; i < (int)(n); i++)  // 0 ~ n-1
#define REPN(i, n) for (int i=1; i <= (int)(n); i++)  // 1 ~ n
#define MIN(vec) min_element(ALL((vec)))  // イテレータのため、値を取り出すときは * を先頭につける
#define MAX(vec) max_element(ALL((vec)))
#define IDX(vec, element_iter) distance((vec).begin(), element_iter)
#define SUM(vec) accumulate(ALL((vec)), 0)  // 0 は初期値
#define COPY(vec1, vec2) copy(ALL(vec1), back_inserter(vec2))  // vec1をvec2にコピーする vec2は空にしておく必要あり
typedef long long ll;

const int MOD = 1000000007;  // 1 000 000 007


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    int N, M, X, Y;
    cin >> N >> M >> X >> Y;
    vector<int> vec;
    // calculation
    vec.push_back(N);
    vec.push_back(M);
    vec.push_back(X);
    vec.push_back(Y);
    sort(vec.begin(), vec.end());
    if (vec[0] == 1 && vec[1] == 4 && vec[2] == 7 && vec[3] == 9) {
        cout << "YES" << "\n";
    } else {
        cout << "NO" << "\n";
    }
    
    
    return 0;
}