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;
#define rep(i,n) for (int (i)=(0);(i)<(int)(n);++(i))
using ll = long long;

const int MAX_N = 1300000;

int prime[MAX_N];
bool is_prime[MAX_N + 1];

void eratos(int n) {
    int p = 0;
    for (int i = 0; i <= n; ++i) is_prime[i] = true;
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i <= n; ++i) {
        if (is_prime[i]) {
            prime[p++] = i;
            for (int j = 2*i; j <= n; j += i) is_prime[j] = false;
        }
    }
}

int main() {
    eratos(MAX_N);

    int n;
    while (cin >> n, n) {

        int r = n, l = n;
        for (int i=n; !is_prime[i]; --i) l=i-1;
        for (int i=n; !is_prime[i]; ++i) r=i+1;

        cout << r-l << endl;

    }
}