// poj 1061 青蛙的约会 拓展欧几里得模板

// 注意进行exgcd时，保证a，b是正数，最后的答案如果是负数，要加上一个膜

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>

#include <cmath>

using namespace std;

typedef long long ll;

ll gcd(ll a,ll b){

    if (b==)

        return a;

    return gcd(b,a%b);

}

ll x,y,m,n,l;

ll exgcd(ll a,ll b,ll &x,ll &y){

    if ( b ==  ){

        x = ;

        y = ;

        return a;

    }

    ll d = exgcd(b,a%b,x,y);

    ll temp = x;

    x = y;

    y = temp - a / b * y;

    return d;

}

int main(){

    //freopen("1.txt","r",stdin);

    while(scanf("%lld%lld%lld%lld%lld",&x,&y,&m,&n,&l)!=EOF){

        ll u = m - n;

        ll v = y - x;

        if (u < ){

            u = -u;

            v = -v;

        }

        ll x1,y1;

        ll d = exgcd(u,l,x1,y1);

        if (v % d){

            puts("Impossible");

            continue;

        }

        ll mod = l / d;

        x1 = x1 * ( v / d);

        printf("%lld\n",(x1%mod + mod)%mod);

    }

    return ;

}