Problem E CodeForces 237C

Description

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 integerx (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.

Sample Input

Input

2 4 2

Output

3

Input

6 13 1

Output

4

Input

1 4 3

Output

-1

一边筛素数，一边处理出一个前缀和sum sum(i)表示[1,i]中有多少素数 那么我们每次查询区间[l,r]中有多少素数，直接查sum[r]-sum[l-1]就可以了 接下去我们按照题意，对答案L进行二分就可以了

#include <iostream>
using namespace std;

const int maxn = 1000001;
int sum[maxn],a,b,k;
bool pri[maxn];
void init(){
    for(int i = 2;i < maxn;i++){
        sum[i] = sum[i-1];
        if(pri[i])  continue;
        sum[i]++;
        for(int j = i+i;j < maxn;j += i)
            pri[j] = 1;
    }
}

bool check(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(){
    init();
    cin.sync_with_stdio(false);
    cin>>a>>b>>k;
    if(sum[b]-sum[a-1] < k){
        cout<<"-1"<<endl;
        return 0;
    }
    int l = 1,r = b-a+1,ans;
    while(l <= r){
        int mid = (l+r)>>1;
        if(check(mid))  ans = mid,r = mid-1;
        else    l = mid+1;
    }
    cout<<ans<<endl;
}