A. Nearly Lucky Number

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”.

問題連結：CodeForces - 110A

問題簡述：判斷一個小於十的十八次方的數即十八個數字中，幸運數字的個數是否為幸運數字，幸運數字為4或7.第一行輸入這串數字，第二行判斷。

問題分析：1、’這串數字為十的十八次方，正好可以用%lld接收這串數字然後每次除十來判斷該數字是否為幸運數字。2、可以用陣列接收十八個數字然後用迴圈判斷該數字是否為幸運數字。3、也可用getchar（）一個數字一個數字地接收並處理。用count統計幸運數字數量並判斷其是否為幸運數字4 7。

程式說明：用getchar（）一個一個接收數字，減少了記憶體佔用，加快了執行速率（？）

AC通過的C語言程式如下：

#include<stdio.h>

int main()
{
	char n,count=0;
	n=getchar();
	while(n!='\n')
	{ 
		if(n=='4'||n=='7')
		{
			count++;
		}
		n=getchar();
	}
	if(count==7||count==4)
	{
		printf("YES\n");
	}else printf("NO\n");
	return 0;
}