CodeForces 110A Nearly Lucky Number-蒲公英云

CodeForces 110A Nearly Lucky Number

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

Copy

output

Copy

NO

input

Copy

output

Copy

YES

input

Copy

1000000000000000000

output

Copy

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”.

题目大意：一个数只包含4 和7称为幸运数字，而一个包含4和7的数的个数为4或7的数称为近幸运数（有点绕）问你一个数是不是幸运数

思路：看懂题意就是大水题……

代码：

/*
*/
#include<map>
#include<set>
#include <vector>
#include<stack>
#include<queue>
#include<cmath>
#include<string>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll unsigned long long
#define inf 0x3f3f3f
#define esp 1e-8
#define bug {printf("mmp\n");}
#define mm(a,b) memset(a,b,sizeof(a))
#define T() int test,q=1;scanf("%d",&test); while(test--)
const int maxn=1e4+10;
const double pi=acos(-1.0);
const int N=201;
const int mod=1e9+7;
char s[maxn];
int a[N];
int main()
{
    int ans=0;
    scanf("%s",s);
    int l=strlen(s);
    for(int i=0;i<l;i++)
    {
        if(s[i]=='4'||s[i]=='7')
            ans++;
    }
    if(ans==4||ans==7)
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}