You've decided to carry out a survey in the theory of prime numbers. Let us remind you that a prime number is a positive integer that has exactly two distinct positive integer divisors.

Consider positive integers a, a + 1, ..., b (a ≤ b). You want to find the minimum integer l (1 ≤ l ≤ b - a + 1) such that for any integer x (a ≤ x ≤ b - l + 1) among lintegers x, x + 1, ..., x + l - 1 there are at least k prime numbers.

Find and print the required minimum l. If no value l meets the described limitations, print -1.

Input

A single line contains three space-separated integers a, b, k (1 ≤ a, b, k ≤ 106; a ≤ b).

Output

In a single line print a single integer — the required minimum l. If there's no solution, print -1.

Examples

Input

2 4 2

Output

3

Input

6 13 1

Output

4

Input

1 4 3

Output

-1

Given a, b is an interval, find a minimum len, so that for all x, x,...,x+len-1, there are at least k prime numbers in this interval.

Among them to be satisfied

a<=x<=b

a<=x-len+1<=b

Dichotomous len, using prefix and optimization.

#define ll long long
const ll mod=1e9+7;
ll n,m;
const ll N = 2000000+999;
ll prime[N] = {0},num_prime = 0;  //prime holds primes less than N
int isNotPrime[N] = {1, 1};        // isNotPrime[i] If I is not a prime, it is 1
int Prime()
{
    for(ll  i = 2 ; i < N ; i ++)
    {
        if(! isNotPrime[i])
            prime[num_prime ++]=i;
        //Whether it's a prime or not, it's going to come down.
        for(ll  j = 0 ; j < num_prime && i * prime[j] <  N ; j ++)
        {
            isNotPrime[i * prime[j]] = 1;
            if( !(i % prime[j] ) )  //Encountering the Minimum Prime Factor of i
                //Key point 1
                break;
        }
    }
    return 0;
}

Preprocessing

Prime();
    for(int i=1; i<=1000006; i++)
    {
        if(!isNotPrime[i])
        {
            s[i]=s[i-1]+1;
        }
        else
        {
            s[i]=s[i-1];
        }
    }

Two points

Dichotomy 1. This version is to find the one with the smallest answer.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1e9+7;
ll n,m;
const ll N = 2000000+999;
ll prime[N] = {0},num_prime = 0;  //prime holds primes less than N
int isNotPrime[N] = {1, 1};        // isNotPrime[i] If I is not a prime, it is 1
int Prime()
{
    for(ll  i = 2 ; i < N ; i ++)
    {
        if(! isNotPrime[i])
            prime[num_prime ++]=i;
        //Whether it's a prime or not, it's going to come down.
        for(ll  j = 0 ; j < num_prime && i * prime[j] <  N ; j ++)
        {
            isNotPrime[i * prime[j]] = 1;
            if( !(i % prime[j] ) )  //Encountering the Minimum Prime Factor of i
                //Key point 1
                break;
        }
    }
    return 0;
}
int s[N];
int a,b,k;
bool check(int len)
{
    for(int i=a; i<=b-len+1; i++)
    {
        if(s[i+len-1]-s[i-1]<k)
            return false;
    }
    return true;
}
int main()
{
    Prime();
    for(int i=1; i<=1000006; i++)
    {
        if(!isNotPrime[i])
        {
            s[i]=s[i-1]+1;
        }
        else
        {
            s[i]=s[i-1];
        }
    }
    scanf("%d %d %d",&a,&b,&k);
    int l=1;
    int r=b-a+1;
    int ans=999999;
    while(l<r)
    {
        int mid=(l+r)/2;//length
        if(check(mid))
        {
            r=mid;
        }
        else
        {
            l=mid+1;
        }
    }
    printf("%d\n",check(l)?l:-1);
}

If the maximum number (>= x) is required:

while(l<r)
    {
        int mid=(l+r+1)/2;//length
        if(a[mid]<=x)
        {
            l=mid;
        }
        else
        {
            r=mid-1;
        }
    }
    return a[l];

Dichotomy 2

This method ensures that there must be an answer to the question.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=1e9+7;
ll n,m;
const ll N = 2000000+999;
ll prime[N] = {0},num_prime = 0;  //prime holds primes less than N
int isNotPrime[N] = {1, 1};        // isNotPrime[i] If I is not a prime, it is 1
int Prime()
{
    for(ll  i = 2 ; i < N ; i ++)
    {
        if(! isNotPrime[i])
            prime[num_prime ++]=i;
        //Whether it's a prime or not, it's going to come down.
        for(ll  j = 0 ; j < num_prime && i * prime[j] <  N ; j ++)
        {
            isNotPrime[i * prime[j]] = 1;
            if( !(i % prime[j] ) )  //Encountering the Minimum Prime Factor of i
                //Key point 1
                break;
        }
    }
    return 0;
}
int s[N];
int a,b,k;
bool check(int len)
{
    for(int i=a; i<=b-len+1; i++)
    {
        if(s[i+len-1]-s[i-1]<k)
            return false;
    }
    return true;
}
int main()
{
    Prime();
    for(int i=1; i<=1000006; i++)
    {
        if(!isNotPrime[i])
        {
            s[i]=s[i-1]+1;
        }
        else
        {
            s[i]=s[i-1];
        }
    }
    scanf("%d %d %d",&a,&b,&k);
    int l=1;
    int r=b-a+1;
    int ans=9999999;
    while(l<=r)
    {
        int mid=(l+r)>>1;//length
        if(check(mid))
        {
            r=mid-1;
            ans=min(ans,mid);
        }
        else
        {
            l=mid+1;
        }
    }
    printf("%d\n",ans==9999999?-1:ans);
}