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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <string>
#include <cstring>
using namespace std;
typedef long long int ll;

const int N=1299709+10;

bool not_prime[N];
vector<int> prime;

void make_prime(){
	not_prime[0]=not_prime[1]=1;
	for(int i=2;i<N;i++){
		if(!not_prime[i]){
			prime.push_back(i);
			for(int j=i+i;j<N;j+=i){
				not_prime[j]=1;
			}
		}
	}
}

int n;

void solve(){
	if(!not_prime[n]){
		printf("0\n");
		return;
	}
	auto it=lower_bound(prime.begin(),prime.end(),n);
	auto it2=it;
	it--;
	printf("%d\n",*it2-*it);
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	make_prime();
	while(1){
		cin >> n;
		if(n==0)break;
		solve();
	}
}
