• 一个需要知道的结论是，比较容易证明
  于是，所以题目就是求的前缀和
  线筛一波，要特判的情况因为，有些头凸

#include<bits/stdc++.h>
#define cs const
using namespace std;
typedef long long ll;
int read(){
  int cnt = 0, f = 1; char ch = 0;
  while(!isdigit(ch)){ ch = getchar(); if(ch == '-') f = -1; }
  while(isdigit(ch)) cnt = cnt*10 + (ch-'0'), ch = getchar();
  return cnt * f;
}
cs int Mod = 1e9 + 7;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
void Add(int &a, int b){ a = add(a, b); }
void Mul(int &a, int b){ a = mul(a, b); }
int sqr(int a){ return mul(a, a); }
cs int N = 1e7 + 50;
int Q, x, A, B, C, Sa, Sb;
int sig[N], sig2[N], mnp[N], vl[N], vl2[N], pc; 
bool isp[N]; int prim[N/5];
void prework(int n){
  sig[1] = sig2[1] = 1;
  for(int i = 2; i <= n; i++){
    if(!isp[i]){
      prim[++pc] = i, 
      vl[i] = sig[i] = 2;
      vl2[i] = sig2[i] = add(sqr(i),1);
      mnp[i] = 1;
    }
    for(int j = 1; j <= pc; j++){
      if(prim[j] * i > n) break;
      int nx = prim[j] * i; isp[nx] = 1;
      if(i % prim[j] == 0){
        mnp[nx] = mnp[i];
        vl[nx] = add(vl[i],1);
        vl2[nx] = add(mul(sqr(prim[j]),vl2[i]),1);
        sig[nx] = mul(sig[mnp[nx]],vl[nx]);
        sig2[nx] = mul(sig2[mnp[nx]],vl2[nx]); 
        break;
      } 
      mnp[nx] = i; 
      vl[nx] = 2; 
      vl2[nx] = add(sqr(prim[j]),1);
      sig[nx] = mul(sig[i],sig[prim[j]]);
      sig2[nx] = mul(sig2[i],sig2[prim[j]]);
    }
  }
}
int main(){
  Q = read();
  x = read(), A = read(), B = read(), C = read();
  prework(C+1);
  while(Q--){
    if(x & 1) Add(Sa, 1), Add(Sb, 4);
    Add(Sa, sig[x]);
    Add(Sb, sig2[x]);
    x = ((ll)x * A + B) % C + 1;
  } cout << Sa << '\n' << Sb; 
  return 0;
}