Description

Two positive integers are said to be relatively prime to each other if the Great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9…are all relatively prime to 2006.

Now your job is easy: for the given integer m, find the K-th element which is relatively prime to m when these elements are sorted in ascending order.

Input

The input contains multiple test cases. For each test case, it contains two integers m (1 <= m <= 1000000), K (1 <= K <= 100000000).

Output

Output the K-th element in a single line.

Sample Input

2006 1

2006 2

2006 3

Sample Output

1

3

5

Solution

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题目大意：给定m和k求 与数m互质的第k个数。

解题思路：gcd(a,b)=gcd(a+k*b,b)

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Code

#include 
     
      #include 
      
       using namespace std;#define N 1000000int prime[N];int gcd(int a, int b){    return b ? gcd(b, a%b) : a;}int main(){    int m, k;    while (~scanf("%d%d", &m, &k))    {        int res = 0;        for (int i = 1; i <=m; i++)            if (gcd(m, i) == 1)                prime[res++] = i;        if (k%res)            cout << k / res*m + prime[k%res - 1] << endl;        else            cout << (k / res - 1)*m + prime[res - 1] << endl;    }    return 0;}