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 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".
解题思路：简单字符串处理。如果输入的字符串中出现'4'或'7'的次数总和num刚好为4次或7次，则输出"YES"。为什么只有4或7呢？因为规定的字符串长度最长为19，而幸运数字（只由4或7组成）有4，7，47...，第三个数47比最长长度值19还大，即num最大为19，所以只需看num是否为4或7，是则输出"YES"，否则输出"NO"。
AC代码：
#include<bits/stdc++.h>
using namespace std;
int main(){
char str[];int num=;
cin>>str;
for(int i=;str[i]!='\0';++i)
if(str[i]=='' || str[i]=='')num++;
if(num== || num==)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
return ;
}