Sohel Sir's Assignment

Sohel sir gave an assignment in CSE-315 course instead of a class test. The assignment was to make questions and provide corresponding answers from the chapters 2, 3, 4, 5. Each student is assigned chapter no y according to the formula:

I.e. he has to make questions and answers from chapter y. According to this rule, Roll 4 was supposed to make questions and answers from chapter 2 as (4 % 4) + 2 = 2 and Roll 35 was assigned to chapter 5 as (35 % 4) +2 = 5. In the meantime, roll 35 had already made the questions & answers from chapter 5 and Roll 4 got the complete assignment of roll 35. So to copy that assignment Roll 4 wanted to change the divisor 4 of the formula to some number m such that his assignment changes to chapter 5 , that is (4 % m) + 2 = 5. But he failed to find such number. Now, your problem is similar to the above problem.

Given two number x and y you have to find a positive number m such that (x % m) + 2 = y. If multiple m is possible, choose the minimum one. If no answer is found print `Impossible'.

Input 

First line of input will contain the number of test cases, T125. Then there follows T lines, each containing two integers x ( 0x10¹²) and y ( 2yx + 2).

Output 

For each case, print m, if m is found. Otherwise print `Impossible' (without quotes). See the samples given below for exact formatting.

Sample Input 

4
4 5
35 5
4 2
11 5

Sample Output 

Impossible
4
1
4

Problem Setter: Mohammad Hafiz Uddin
Alternate Solution: Radi Muhammad Reza

---

x%m+2  = y
x%m = y-2
x = km + y-2
x-y+2 = km
分解 x-y+2 即可，如果 x-y+2 == 0，輸出 x+1。

#include <stdio.h>
#include <math.h>
int main() {
    scanf("%*d");
    long long x, y;
    int i;
    while(scanf("%lld %lld", &x, &y) == 2) {
        long long n = x-(y-2), sq;
        sq = (long long)sqrt(n);
        if(n == 0) {
            printf("%lld\n", x+1);
            continue;
        }
        int flag = 0;
        long long mn;
        for(i = 1; i <= sq; i++) {
            if(n%i == 0) {
                if(x%i+2 == y) {
                    if(!flag)   mn = i, flag = 1;
                    if(i < mn)  mn = i;
                }
                if(x%(n/i)+2 == y) {
                    if(!flag)   mn = n/i, flag = 1;
                    if(n/i < mn)  mn = n/i;
                }
            }
            if(flag && i > mn)  break;
        }
        if(flag)
            printf("%lld\n", mn);
        else
            puts("Impossible");
    }
    return 0;
}

---