#include <stdio.h>
#include <math.h>
int b[999999];

int prime(int x) {
	for (int i = 2; i < sqrt(x) + 1; i++) {
		if (x%i == 0) {
			return 0;
		}
	}
	return 1;
}
int main(){
	int n;
	b[0] = 0;
	b[1] = 0;
	b[2] = 1;
	int count = 1;
	for (int i = 3; i <= 999999; i++) {
		if (prime(i))count++;
		b[i] = count;
	}
	while (scanf("%d", &n) != EOF){
		printf("%d\n", b[n]);
	}
	return 0;
}
### Prompt

Develop a solution in Cpp 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>
#include <math.h>
int b[999999];

int prime(int x) {
	for (int i = 2; i < sqrt(x) + 1; i++) {
		if (x%i == 0) {
			return 0;
		}
	}
	return 1;
}
int main(){
	int n;
	b[0] = 0;
	b[1] = 0;
	b[2] = 1;
	int count = 1;
	for (int i = 3; i <= 999999; i++) {
		if (prime(i))count++;
		b[i] = count;
	}
	while (scanf("%d", &n) != EOF){
		printf("%d\n", b[n]);
	}
	return 0;
}
```