#include <iostream>
#include <algorithm>

using namespace std;

typedef long long int lli;
const lli mx = 1299711;

int main()
{
  lli n, s, t;
  bool a[mx];
  fill(a, a+mx, true);
  a[0] = a[1] = false;

  for(lli i = 2; i < mx; i++){
    if(a[i]){
      for(lli j = i+i; j < mx; j += i){
	a[j] = false;
      }
    }
  }

  while(cin>>n && n){
    for(lli i = n; i >= 2; i--){
      if(a[i]){
	s = i;
	break;
      }
    }
    for(lli i = n; i < mx; i++){
      if(a[i]){
	t = i;
	break;
      }
    }

    cout << (t-s) << endl;
  }

  return 0;
}
### Prompt

Please formulate a cpp solution to the following problem:

The sequence of n - 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive prime numbers p and p + n is called a prime gap of length n. For example, (24, 25, 26, 27, 28) between 23 and 29 is a prime gap of length 6.

Your mission is to write a program to calculate, for a given positive integer k, the length of the prime gap that contains k. For convenience, the length is considered 0 in case no prime gap contains k.



Input

The input is a sequence of lines each of which contains a single positive integer. Each positive integer is greater than 1 and less than or equal to the 100000th prime number, which is 1299709. The end of the input is indicated by a line containing a single zero.

Output

The output should be composed of lines each of which contains a single non-negative integer. It is the length of the prime gap that contains the corresponding positive integer in the input if it is a composite number, or 0 otherwise. No other characters should occur in the output.

Example

Input

10
11
27
2
492170
0


Output

4
0
6
0
114

### Response

```cpp
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long int lli;
const lli mx = 1299711;

int main()
{
  lli n, s, t;
  bool a[mx];
  fill(a, a+mx, true);
  a[0] = a[1] = false;

  for(lli i = 2; i < mx; i++){
    if(a[i]){
      for(lli j = i+i; j < mx; j += i){
	a[j] = false;
      }
    }
  }

  while(cin>>n && n){
    for(lli i = n; i >= 2; i--){
      if(a[i]){
	s = i;
	break;
      }
    }
    for(lli i = n; i < mx; i++){
      if(a[i]){
	t = i;
	break;
      }
    }

    cout << (t-s) << endl;
  }

  return 0;
}
```