CodeForces

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 l integers 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.

Example

Input

2 4 2

Output

3

Input

6 13 1

Output

4

Input

1 4 3

Output

-1

题意：找到最小的l，使[a,b]中所有长度为l的区间都能找到k个素数，如果找不到，输出"-1"
分析：   答案是线性相关的，所以我们可以二分答案，然后再通过尺取法判断该长度是否满足条件
代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int MAXN=2e6+10;
int a,b,k;
int flag[MAXN];
int flag2;
int que[MAXN];
bool check(int len)
{
 //   puts("c");
    int num,len1,l,r,now;
    int flag3;
    flag3=1;
    l=0,r=0;
    now=0;
    while(1)
    {
         while(r-l<len)
         {
          que[r++]=a+r-1;
          if(flag[a+r-1]){
          now++;
          }
          if(a+r-1==b)break;
         }
        if(r-l==len)
        {
          if(now<k)
          flag3=0;
        //  cout<<"    "<<que[l]<<endl;
          if(flag[que[l]]){
          now--;
          }
          l++;
        }
        if(flag3==0||a+r-1==b)
        break;
    }
    if(flag3==1)return true;
    return false;
}
int main()
{
    flag[1]=0;
    flag[2]=1;
    for(int i=3;i<=1000100;i++)
    {
       flag2=1;
       for(int j=2;j*j<=i;j++)
       {
           if(i%j==0)
           {
               flag2=0;
               break;
           }
       }
       flag[i]=flag2;
    }
    while(scanf("%d%d%d",&a,&b,&k)!=EOF)
    {
        int l,r,mid;
        l=0; r=b-a+1;
        while(l+1<r)
        {
          mid=(l+r)/2;
          if(check(mid))
          r=mid;
          else
          l=mid;
        }
       if(check(r))printf("%d
",r);
       else puts("-1");
    }
    return 0;
}