#include<stdio.h>
int prime[1000000];
int main(){
    prime[0]=-1;
    prime[1]=-1;
    for(int i=2;i<1000000;i++){
        if(prime[i]==0){
            prime[i]=1;
            for(int j=i*2;j<1000000;j+=i)prime[j]=-1;
        }
    }
    int now;
    while(~scanf("%d",&now)){
        int ans=0;
        for(int i=1;i<now+1;i++)if(prime[i]==1)ans++;
        printf("%d\n",ans);
    }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

Write a program which reads an integer n and prints the number of prime numbers which are less than or equal to n. A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.



Input

Input consists of several datasets. Each dataset has an integer n (1 ≤ n ≤ 999,999) in a line.

The number of datasets is less than or equal to 30.

Output

For each dataset, prints the number of prime numbers.

Example

Input

10
3
11


Output

4
2
5

### Response

```cpp
#include<stdio.h>
int prime[1000000];
int main(){
    prime[0]=-1;
    prime[1]=-1;
    for(int i=2;i<1000000;i++){
        if(prime[i]==0){
            prime[i]=1;
            for(int j=i*2;j<1000000;j+=i)prime[j]=-1;
        }
    }
    int now;
    while(~scanf("%d",&now)){
        int ans=0;
        for(int i=1;i<now+1;i++)if(prime[i]==1)ans++;
        printf("%d\n",ans);
    }
}

```