Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number n is a nearly lucky number.

Input

The only line contains an integer n (1 ≤ n ≤ 1018).

Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.

Output

Print on the single line "YES" if n is a nearly lucky number. Otherwise, print "NO" (without the quotes).

Examples

Input

40047


Output

NO


Input

7747774


Output

YES


Input

1000000000000000000


Output

NO

Note

In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".

In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".

In the third sample there are no lucky digits, so the answer is "NO".
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main{
	
	public static void main(String[] args) throws IOException {
		
		FastScanner sc = new FastScanner(System.in);
		
		long t = sc.nextLong();
		int c = 0;
		
		while (t > 0) {
			long last = t % 10;
			
			if (last == 4 || last == 7) {
				c++;
			}
			
			t /= 10;
		}
		
		if (c == 4 || c == 7)
			System.out.println("YES");
		else
			System.out.println("NO");
	}
	
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new
                    InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
        	return Long.parseLong(next());
        }
    }
}
