題目描述：

Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every m-th day (at days with numbers m, 2m, 3m, ...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?

Input

The single line contains two integers n and m (1 ≤ n ≤ 100; 2 ≤ m ≤ 100), separated by a space.

Output

Print a single integer — the answer to the problem.

Examples

Input

2 2

Output

3

Input

9 3

Output

13

Note

In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.

In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.

題目大意：輸入兩個數字，第一個代表Vasya現有的襪子，第二個數代表每隔幾天媽媽會給Vasya買一雙新襪子，問Vasya的襪子最多有多少天可以穿襪子。

程式碼：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
const int maxn=500000;
int main()
{
   int n,m;
   while(scanf("%d%d",&n,&m)!=EOF)
   {
       int sum=0;
       int count1=n;
       for(int i=1;i<=n;i++)
       {
           sum++;
           if(sum==m)//媽媽買襪子的那一天
           {
               count1++;襪子數加1
               i--;//迴圈最多到n,增加了一雙襪子可以多迴圈一天了，所以i--
               sum=0;//讓sum為0，繼續迴圈的媽媽買襪子的那天
           }

       }
       printf("%d\n",count1);
   }
   return 0;
}