R.T.

//#define DEBUG
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
#include <cctype>
#include <cstdlib>
#include <utility>
#include <queue>
#include <stack>
#include <deque>
#include <iomanip>
#include <vector>
#include <list>
#include <set>
using namespace std;
using ll=long long ;
ll Exgcd(ll a,ll b,ll &p,ll &q)
{
	if(!b) {
		p=1,q=0;
		return a;
	}
	
	ll ret=Exgcd(b,a%b,p,q);
	ll tmp=p;
	p=q,q=tmp-a/b*q;
	return ret;
}

int main()
{
	ll x,y,m,n,L,p,q;
	scanf("%lld%lld%lld%lld%lld",&x,&y,&m,&n,&L);
	
	ll gcd=Exgcd(m-n,L,p,q);
	if((y-x)%gcd) {
		printf("Impossible");
		return 0;
	}
	/**/
	ll k=(y-x)/gcd;
	p*=k,q*=k;ll t=L/gcd;
#ifdef DEBUG
	printf("a:%lld,b:%lld,c:%lld,gcd:%lld\n",m-n,L,y-x,gcd);
	printf("p:%lld,q:%lld,t:%lld\n",p,q,t);
#endif
	
	/*?不理解这一句，请Dalao改正后解释一下呗~~~*/
	p=(p%L+L)%L;
	
	printf("%lld",p);
	return 0;
}