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;

int prime[100010];
bool p[1299710];

void eratos(){
    fill(p, p+1299710, true);
    p[0]=p[1]=false;
    for(int i=2; i*i<=1299709; i++){
        if(p[i]){
            for(int j=i; i*j<=1299709; j++){
                p[i*j]=false;
            }
        }
    }
    int now=0;
    for(int i=2; i<=1299709; i++){
        if(p[i]){
            prime[now++]=i;
        }
    }
}

int main(){
    int n;
    eratos();
    
    while(cin>>n, n){
        if(p[n]) {
            cout<<0<<"\n";
        }
        else{
            int idx=(int)(lower_bound(prime, prime+100001, n)-prime);
            cout << prime[idx]-prime[idx-1]<<"\n";
        }
    }
    
}
