#include <iostream>
#include <vector>

using namespace std;

const int N = 1299710;

vector<bool> p;

void init(){
    p.assign(N, true); p[0] = p[1] = false;
    for(int i=2; i<=N; ++i)if(p[i]){
        for(int j=i<<1; j<=N; j+=i)p[j] = false;
    }
}

int main(){
    init();
    for(int k; cin >> k, k;){
        int lo = k, hi = k;
        while(!p[lo])lo--;
        while(!p[hi])hi++;
        cout << hi - lo << '\n';
    }
    return 0;
}
### 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>
#include <vector>

using namespace std;

const int N = 1299710;

vector<bool> p;

void init(){
    p.assign(N, true); p[0] = p[1] = false;
    for(int i=2; i<=N; ++i)if(p[i]){
        for(int j=i<<1; j<=N; j+=i)p[j] = false;
    }
}

int main(){
    init();
    for(int k; cin >> k, k;){
        int lo = k, hi = k;
        while(!p[lo])lo--;
        while(!p[hi])hi++;
        cout << hi - lo << '\n';
    }
    return 0;
}
```