推式子 （用自己求自己）（ % 构造新的自变量 数论么？，先乘后除（%） 要小心 ）

Vasya likes to solve equations. Today he wants to solve (x div k)⋅(xmodk)=n, where div and mod stand for integer division and modulo operations (refer to the Notes below for exact definition). In this equation, k and n are positive integer parameters, and x is a positive integer unknown. If there are several solutions, Vasya wants to find the smallest possible x. Can you help him?

Input
The first line contains two integers n and k (1≤n≤106, 2≤k≤1000).

Output
Print a single integer x — the smallest positive integer solution to (x div k)⋅(xmodk)=n. It is guaranteed that this equation has at least one positive integer solution.

Examples
Input
6 3
Output
11
Input
1 2
Output
3
Input
4 6
Output
10
Note
The result of integer division a div b is equal to the largest integer c such that b⋅c≤a. a modulo b (shortened amodb) is the only integer c such that 0≤c<b, and a−c is divisible by b.

In the first sample, 11 div 3=3 and 11mod3=2. Since 3⋅2=6, then x=11 is a solution to (x div 3)⋅(xmod3)=6. One can see that 19 is the only other positive integer solution, hence 11 is the smallest one.

//注意先乘后除会wa的！！！！！！！

#pragma warning(disable:4996)
#include"iostream"
#include"functional"
#include"algorithm"
#include"cstring"
#include"stack"
#include"cmath"
#include"queue"
#include"vector"
#include"map"
typedef long long int ll;
using namespace std;
int main() {
	ll a, b;
	ll mi =0;
	cin >> a >> b;
	ll i = 1;
	for (; i <= b - 1; i++) {
		if ((a*b) % i == 0) {
			mi =  (a / i* b)  + i;
			i++;
			break;
		}
	}
	for (; i <= b - 1; i++) {
		if ((a*b)% i == 0) {
			mi = mi > a/ i* b  + i ? a/ i * b  + i : mi;
		}
	}
	cout << mi<<endl;
}