CodeForces 237C Primes on Interval

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

题意：给你一个闭区间[a,b]，求一个最小的L，使得在区间[a,b-L+1]内任取一个数x，可以满足在x,x+1,x+2,……,x+L-2,x+L-1内至少包含k个素数。(1<=a,b,k<=10^6)

思路：素数打表+二分

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;bool prim[1100000];int sum[1100000]; int a,b,k;void getprim()//素数打表，并求 1到i区间内素数的个数{int i,j;for(i=2;i<=1000000;i++){sum[i]=sum[i-1];if(prim[i])continue;sum[i]++;for(j=i+i;j<=1000000;j+=i)prim[j]=1;}}int judge(int mid){for(int i=a;i<=b-mid+1;i++){if(sum[i+mid-1]-sum[i-1]<k) return 0;}return 1;}int main(){getprim();while(scanf("%d%d%d",&a,&b,&k)!=EOF){if(sum[b]-sum[a-1]<k){printf("-1\n");continue;}int l,r,mid,ans;l=1;r=b-a+1;while(l<=r){mid=(l+r)/2;if(judge(mid)){ans=mid;r=mid-1;}else{l=mid+1;}}printf("%d\n",ans);}return 0;}