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
#include<bits/stdc++.h>
using namespace std;

bool isprime[1299710];

void eratos(int n){
  
  for(int i=0 ; i <= n ; i++){
    isprime[i]=true;    
  }
  isprime[0]=isprime[1]=false;
  for(int i=2 ; i <= sqrt(n) ; i++){
    if(isprime[i]){
      int  j=i+i;
      while(j <= n){
	isprime[j]=false;
	j=j+i;
      }
    }
  }
}

int main(){
  
  int k;
  int prime1,prime2;

  eratos(1299709);

  while(1){
    
    cin >>k;
    if(k==0)break;
    
    if(isprime[k]){
      cout <<"0"<<endl;
    }else{
      for(prime1=k+1 ; !isprime[prime1] ; prime1++);
      for(prime2=k-1 ; !isprime[prime2] ; prime2--);
      cout <<prime1-prime2<<endl;;
    }
 
  }
 
  return 0;
}