Problem description

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

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

40047

NO

7747774

YES

1000000000000000000

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".

解題思路：簡單字符串處理。如果輸入的字符串中出現‘4‘或‘7‘的次數總和num剛好為4次或7次，則輸出"YES"。為什麽只有4或7呢？因為規定的字符串長度最長為19，而幸運數字（只由4或7組成）有4，7，47...，第三個數47比最長長度值19還大，即num最大為19，所以只需看num是否為4或7，是則輸出"YES"，否則輸出"NO"。

AC代碼：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     char str[20];int num=0;
 5     cin>>str;
 6     for(int i=0;str[i]!=‘\0‘;++i)
 7         if(str[i]==‘4‘ || str[i]==‘7‘)num++;
 8     if(num==4 || num==7)cout<<"YES"<<endl;
 9     else cout<<"NO"<<endl;
10     return 0;
11 }

B - Nearly Lucky Number