【Codeforcdes 237C. Primes on Interval】+ 二分

C. Primes on Interval

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

Input

2 4 2

Output

3

Input

6 13 1

Output

4

Input

1 4 3

Output

-1

二分~注意姿势即可·~~

AC代码：

#include<cstdio>
const int MAXN = 1e6 + 10;
int num[MAXN],a,b,k;
bool solve(int x){
for(int i = a ; i <= b - x + 1; i++)
if(num[i + x - 1] - num[i - 1] < k)
return false;
return true;
}
int main()
{
int cut = 0;
for(int  i = 2 ; i <= 1e6 ; i++){
int ok = 1;
for(int j = 2 ; j * j <= i ; j++)
if(i % j == 0){
ok = 0;
break;
}
if(ok) cut++;
num[i] = cut;
}
scanf("%d %d %d",&a,&b,&k);
int low = 1,high = 1e6,ans = -1;
while(low <= high){
int mid = (low + high) / 2;
if(solve(mid)){
if(mid <= b - a + 1 && mid >= 1)
ans = mid;
high = mid - 1;
}
else low = mid + 1;
}
printf("%d\n",ans);
return 0;
}