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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)



import java.io.*;
import java.text.*;
import java.util.*;

public class Main {

    static InputReader sc = new InputReader(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) throws IOException {
        solver s = new solver();
        int t = 1;
        while (t > 0) {
            s.solve();
            t--;
        }
        out.close();
    }

    static class InputReader {

        public BufferedReader br;
        public StringTokenizer token;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            token = null;
        }

        public String sn() {
            while (token == null || !token.hasMoreTokens()) {
                try {
                    token = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }

        public int ni() {
            return Integer.parseInt(sn());
        }

        public String snl() throws IOException {
            return br.readLine();
        }

        public long nlo() {
            return Long.parseLong(sn());
        }

        public double nd() {
            return Double.parseDouble(sn());
        }

        public int[] na(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.ni();
            return a;
        }

        public long[] nal(int n) {
            long a[] = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = sc.nlo();
            return a;
        }
    }

    static int inf = 5000013;

    static class solver {
        DecimalFormat df = new DecimalFormat("0.00000000000000000000");
        long max = 1000000000000000l, min = -10000000000000000l;
        long mod = (long) 1e9 + 7;

        long pro(long a, long b) {
            return (a % mod * b % mod) % mod;
        }

        long add(long a, long b) {
            return (a + b) % mod;
        }

        long sub(long a, long b) {
            return ((a % mod - b % mod) + mod) % mod;
        }

        long ncr3(long n) {
            return (n * (n - 1) * (n - 2)) / 6l;
        }

        long ncr2(long n) {
            return n * (n - 1) / 2l;
        }

        long ncr(long n, long r) {
            double res = 1.0;
            for (long i = n, j = r; i > (n - r); i--, j--) {
                res *= (i * 1.0) / j;
            }
            return (long) Math.ceil(res);
        }

        long modPow(long x, long y) {
            long res = 1;
            x = x % mod;
            while (y > 0) {
                if (y % 2 != 0)
                    res = pro(res, x);
                x = pro(x, x);
                y = y >> 1;
            }
            return res;
        }

        int gcd(int a, int b) {
            if (b == 0)
                return a;
            return gcd(b, a % b);
        }

        final int maxn = 5005;

        void solve() throws IOException {
            int a, b, c;
            a = sc.ni();
            b = sc.ni();
            c = sc.ni();
            boolean ans = false;
            int d[][] = {{5, 5, 7}, {5, 7, 5}, {7, 5, 5}};
            for (int i = 0; i < 3; i++) {
                if (a == d[i][0] && b == d[i][1] && c == d[i][2]) {
                    ans = true;
                }
            }
            if (ans)
                out.println("YES");
            else
                out.println("NO");
        }
    }
}