『NYIST』第九届河南省ACM竞赛队伍选拔赛[正式赛二]- Nearly Lucky Number（Codeforces Beta Round #84 (Div. 2 Only)A. Nearly）

A. Nearly Lucky Number

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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".

很幸运看到了这道水题，立马把它A了，题意就是一个数所包含4或者7的数量如果是Lucky Number，那么就输出YES,反之，NO,水吧，重要在样例，一般看hint或note,这是最关键的；

<pre name="code" class="plain">#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=50;
char a[N];
int main()
{
    int x,i,j;
    while(~scanf("%s",a))
    {
        i=j=0;
        x=strlen(a);
        while(x--)
        {
            if(a[i]=='4'||a[i]=='7')
                j++;
            i++;
        }
        int f=0;
        if(j==0)
                                            f=1;                            

                                         else
        {
            while(j)//其实这可以不这样写的，数据范围并不大，j如果符合肯定是个位数；           {只需判断j是否等于4或7；
                if(j%10!=4&&j%10!=7)
                    f=1;
                j/=10;
            }
        }

        if(f)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}