#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll=long long;
using ld=long double;
ld pie=3.141592653589793;
ll inf=155550555555555;
ll mod=1000000007;
ll modpow(ll x, ll n) {
  x = x%mod;
  if(n==0) return 1;  //再帰の終了条件

  else if(n%2==1) {
    return (x*modpow(x, n-1))%mod;  //nが奇数ならnを1ずらす
  }
  else return modpow((x*x)%mod, n/2)%mod;  //nが偶数ならnが半分になる
}
int main(){
    ll n;
    cin >> n;
    ll a,b,c;
    cin >> a >> b >> c;
    n-=2;
    vector<tuple<ll,ll,ll>>t(6);
    t[0]={a-b,b-c,c-a};
    t[1]={a-2*b+c,a+b-2*c,-2*a+b+c};
    t[2]={-3*b+3*c,3*a-3*c,3*b-3*a};
    t[3]={-3*a-3*b+6*c,6*a-3*b-3*c,-3*a+6*b-3*c};
    t[4]={-9*a+9*c,9*a-9*b,9*b-9*c};
    t[5]={-18*a+9*b+9*c,9*a-18*b+9*c,9*a+9*b-18*c};
    for (ll i = 0; i < t.size(); i++)
    {
        get<0>(t[i])%=mod;
        get<1>(t[i])%=mod;
        get<2>(t[i])%=mod;
    }
    ll z=n/6;
    ll x=modpow(-27,z);
    x+=mod;
    x%=mod;
    n%=6;
    z=n;
    ll aa=(get<0>(t[z])*x)%mod,bb=(get<1>(t[z])*x)%mod ,cc=(get<2>(t[z])*x)%mod;
    aa+=mod;
    aa%=mod;
    bb+=mod;
    bb%=mod;
    cc+=mod;
    cc%=mod;
    cout << aa << ' ' << bb << ' ' << cc << endl;
}