Crashing Balloon

On every June 1st, the Children‘s Day, there will be a game named "crashing balloon" on TV.   The rule is very simple.  On the ground there are 100 labeled balloons, with the numbers 1 to 100.  After the referee shouts "Let‘s go!" the two players, who each starts with a score of  "1", race to crash the balloons by their feet and, at the same time, multiply their scores by the numbers written on the balloons they crash.  After a minute, the little audiences are allowed to take the remaining balloons away, and each contestant reports his\her score, the product of the numbers on the balloons he\she‘s crashed.  The unofficial winner is the player who announced the highest score.

Inevitably, though, disputes arise, and so the official winner is not determined until the disputes are resolved.  The player who claims the lower score is entitled to challenge his\her opponent‘s score.  The player with the lower score is presumed to have told the truth, because if he\she were to lie about his\her score, he\she would surely come up with a bigger better lie.  The challenge is upheld if the player with the higher score has a score that cannot be achieved with balloons not crashed by the challenging player.  So, if the challenge is successful, the player claiming the lower score wins.

So, for example, if one player claims 343 points and the other claims 49, then clearly the first player is lying; the only way to score 343 is by crashing balloons labeled 7 and 49, and the only way to score 49 is by crashing a balloon labeled 49.  Since each of two scores requires crashing the balloon labeled 49, the one claiming 343 points is presumed to be lying.

On the other hand, if one player claims 162 points and the other claims 81, it is possible for both to be telling the truth (e.g. one crashes balloons 2, 3 and 27, while the other crashes balloon 81), so the challenge would not be upheld.

By the way, if the challenger made a mistake on calculating his/her score, then the challenge would not be upheld. For example, if one player claims 10001 points and the other claims 10003, then clearly none of them are telling the truth. In this case, the challenge would not be upheld.

Unfortunately, anyone who is willing to referee a game of crashing balloon is likely to get over-excited in the hot atmosphere that he\she could not reasonably be expected to perform the intricate calculations that refereeing requires.  Hence the need for you, sober programmer, to provide a software solution.

Input

Pairs of unequal, positive numbers, with each pair on a single line, that are claimed scores from a game of crashing balloon.

Output

Numbers, one to a line, that are the winning scores, assuming that the player with the lower score always challenges the outcome.

Sample Input

343 49
3599 610
62 36

Sample Output

49
610
62

这道题题目有点难理解。

判定标准是： b不服，站出来质疑；
如果a能举证说：你瞧，存在一种合理的解释，a = a[1]*a[2]*…*a[n], b = b[1]*b[2]*…*b[m];
其中 2 <= a[i], b[j] <= 100, 且 a[i] !=b[j] if i!=j 就判断a赢，否则b赢
但是，这里会有对b不利的冤案！例如b踩了气球4和8，于是b = 32 而 a吹牛说自己得了44分，
b不服，但是a狡猾的说，你看，a = 4*11, b=2*16，于是a赢了！b虽然说了实话，a说了假话，
但是仍然b输了。所以我觉得，这道题目应该是按照：谁主张，谁举证的原则来判决。
既然b主张，那么此时法官应该问b，你踩了哪些气球，b当然可以诬告，但是b此时必须说出一种分解，如果b不能分解，
那么显然是诬告，法官不用问a，就判断a赢；
但是针对b说的**任意(arbitrary)**分解，如果a不能给出一种合理的解释，此时法官就可以判断b赢，否则判断a赢。

/*
ID: LinKArftc
PROG: 1003.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

int fa, fb;

void dfs(int a, int b, int k) {
    if (b == 1) {
        fb = 1;
        if (a == 1) fa = 1;
    }
    if (k == 101 || (fa && fb)) return;
    if (a % k == 0) dfs(a / k, b, k + 1);
    if (b % k == 0) dfs(a, b / k, k + 1);
    dfs(a, b, k + 1);
}

int main() {

    int a, b;
    while (~scanf("%d %d", &a, &b)) {
        if (a < b) { fa = a; a = b; b = fa; }
        fa = fb = 0;
        dfs(a, b, 2);
        if (fb == 1) {
            if (fa == 1) printf("%d\n", a);
            else printf("%d\n", b);
        } else printf("%d\n", a);
    }

    return 0;
}