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

vector<int> prime;
vector<int> ls;

void sieve(int N){
    prime.resize(N+1,1);
    prime[0] = prime[1] = 0;
    for(int i=2;i*i<=N;i++){
        for(int j=i*i;j<=N;j+=i){
            prime[j] = 0;
        }
    }
    for(int i=2;i<=N;i++){
        if(prime[i])ls.push_back(i);
    }
}

int main() {
    sieve(int(2e6));
    int n;
    while(cin >> n, n){
        int l=n, r=n;
        while(prime[l]==0)l--;
        while(prime[r]==0)r++;
        cout << r-l << endl;
    }
    return 0;
}

### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;

vector<int> prime;
vector<int> ls;

void sieve(int N){
    prime.resize(N+1,1);
    prime[0] = prime[1] = 0;
    for(int i=2;i*i<=N;i++){
        for(int j=i*i;j<=N;j+=i){
            prime[j] = 0;
        }
    }
    for(int i=2;i<=N;i++){
        if(prime[i])ls.push_back(i);
    }
}

int main() {
    sieve(int(2e6));
    int n;
    while(cin >> n, n){
        int l=n, r=n;
        while(prime[l]==0)l--;
        while(prime[r]==0)r++;
        cout << r-l << endl;
    }
    return 0;
}

```