#include <iostream>
using namespace std;

const int MAX_N=1299709;
bool prime[MAX_N+1];

void sieve()
{
  prime[0]=prime[1]=false;
  for(int i=2;i<=MAX_N;i++)
    prime[i]=true;
  for(int i=2;i<=MAX_N;i++)
    if(prime[i])
      for(int j=i*2;j<=MAX_N;j+=i)
	prime[j]=false;
}

int is_prime(int n)
{
  return prime[n];
}

int main()
{
  int n,sum;

  sieve();
  for(;;){
    cin>>n;
    if(!n) break;

    sum=0;
    for(int i=n;;i++){
      if(is_prime(i)) break;
      sum++;
    }
    for(int i=n;;i--){
      if(is_prime(i)) break;
      sum++;
    }
    cout<<sum<<endl;
  }
}

### Prompt

Your task is to create 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>
using namespace std;

const int MAX_N=1299709;
bool prime[MAX_N+1];

void sieve()
{
  prime[0]=prime[1]=false;
  for(int i=2;i<=MAX_N;i++)
    prime[i]=true;
  for(int i=2;i<=MAX_N;i++)
    if(prime[i])
      for(int j=i*2;j<=MAX_N;j+=i)
	prime[j]=false;
}

int is_prime(int n)
{
  return prime[n];
}

int main()
{
  int n,sum;

  sieve();
  for(;;){
    cin>>n;
    if(!n) break;

    sum=0;
    for(int i=n;;i++){
      if(is_prime(i)) break;
      sum++;
    }
    for(int i=n;;i--){
      if(is_prime(i)) break;
      sum++;
    }
    cout<<sum<<endl;
  }
}

```