#include <bits/stdc++.h>

using namespace std;
using ll = long long;
 
#define rep(i, n)      for (int i = 0; i < (n); i++)
#define repr(i, n)     for (int i = (n) - 1; i >= 0; i--)
#define repe(i, l, r)  for (int i = (l); i < (r); i++)
#define reper(i, l, r) for (int i = (r) - 1; i >= (l); i--)
#define repi(i, l, r)  for (int i = (l); i <= (r); i++)
#define repir(i, l, r) for (int i = (r); i >= (l); i--)
#define range(a) a.begin(), a.end()
void initio() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); }

ll MOD;

class mint {
  ll n;
public:
  mint(ll n_ = 0) : n(n_) {}
  explicit operator ll() { return n; }
  friend mint operator-(mint a) { return -a.n + MOD * (a.n != 0); }
  friend mint operator+(mint a, mint b) { ll x = a.n + b.n; return x - (x >= MOD) * MOD; }
  friend mint operator-(mint a, mint b) { ll x = a.n - b.n; return x + (x < 0) * MOD; }
  friend mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; }
  friend mint &operator+=(mint &a, mint b) { return a = a + b; }
  friend mint &operator-=(mint &a, mint b) { return a = a - b; }
  friend mint &operator*=(mint &a, mint b) { return a = a * b; }
  friend bool operator==(mint a, mint b) { return a.n == b.n; }
  friend bool operator<(mint a, mint b) { return a.n < b.n; }
  friend bool operator!=(mint a, mint b) { return a.n != b.n; }
  friend istream &operator>>(istream &i, mint &a) { return i >> a.n; }
  friend ostream &operator<<(ostream &o, mint a) { return o << a.n; }
};

mint modpow(mint a, ll b) {
  mint res = 1;
  while (b > 0) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}

mint modinv(mint n) {
  ll a = (ll)n, b = MOD;
  ll s = 1, t = 0;
  while (b != 0) {
    int q = a / b;
    a -= q * b;
    s -= q * t;
    swap(a, b);
    swap(s, t);
  }
  return s >= 0 ? s : s + MOD;
}


using mat = vector<mint>;
mat E = {1, 0, 0, 1};

mint det(mat A) {
  return A[0] * A[3] - A[1] * A[2];
}

mat inv(mat A) {
  mint d = modinv(det(A));
  mat res(4);
  res[0] = A[3] * d;
  res[1] = -A[1] * d;
  res[2] = -A[2] * d;
  res[3] = A[0] * d;
  return res;
}

mat mul(mat A, mat B) {
  mat res(4);
  res[0] = A[0] * B[0] + A[1] * B[2];
  res[1] = A[0] * B[1] + A[1] * B[3];
  res[2] = A[2] * B[0] + A[3] * B[2];
  res[3] = A[2] * B[1] + A[3] * B[3];
  return res;
}

mat matpow(mat A, ll B) {
  mat res = E;
  while (B > 0) {
    if (B & 1) res = mul(res, A);
    A = mul(A, A);
    B >>= 1;
  }
  return res;
}

ll bsgs(mint a, mint b) {
  constexpr ll S = 70000;
  map<mint, ll> mp;
  mint R = 1;
  for (int i = 0; i < S; i++) {
    if (!mp.count(R)) {
      mp[R] = i;
    }
    R *= a;
  }
  // A^{50000i + j} = B
  // A^j = B^A{-50000i}
  R = modinv(R);
  for (int i = 0; i < S; i++) {
    if (mp.count(b)) {
      return S*i + mp[b];
    }
    b *= R;
  }
  return -1;
}

ll bsgs_except_zero(mint a, mint b) {
  constexpr ll S = 70000;
  map<mint, ll> mp;
  mint R = 1;
  for (int i = 0; i < S; i++) {
    if (!mp.count(R * a)) {
      mp[R * a] = i + 1;
    }
    R *= a;
  }
  // A^{50000i + j} = B
  // A^j = B^A{-50000i}
  R = modinv(R);
  for (int i = 0; i < S; i++) {
    if (mp.count(b)) {
      return S*i + mp[b];
    }
    b *= R;
  }
  return -1;
}

ll bsgs_mat(mat A, mat B) {
  constexpr ll S = 70000;
  map<mat, ll> mp;
  mat R = E;
  for (int i = 0; i < S; i++) {
    if (!mp.count(R)) {
      mp[R] = i;
    }
    R = mul(R, A);
  }
  // A^{50000i + j} = B
  // A^j = B^A{-50000i}
  R = inv(R);
  for (int i = 0; i < S; i++) {
    if (mp.count(B)) {
      return S*i + mp[B];
    }
    B = mul(B, R);
  }
  return -1;
}

// ----------------------------------------------------------------
// det(A)=0 のときケーリーハミルトンの定理より A^2-(a+d)A = O が成り立つ。つまり
// A^n = (a+d)^{n-1} A = B
// これは BSGS で解ける。
// ----------------------------------------------------------------
// det(A)!=0 のとき。
// 
// det(A)^N = det(B) より N%T が求められる。ここで T は det(A) の位数とする。
// よって
// A ^ {N%T + Tk} = B
// det(A) != 0 なので A に逆行列があり
// (A^T)^k = B^A^{-N%T}
// 
// det(A^T)=1 なので A^T の周期は 2p 以下。これで BSGS が使える。

int main() { initio();
  cin >> MOD;
  mat A(4), B(4);
  rep(i, 4) cin >> A[i];
  rep(i, 4) cin >> B[i];
  if (det(A) == 0) {
    int k = -1;
    rep(i, 4) if (A[i] != 0) k = i;
    if (k == -1) {
      if (A == B) {
        cout << 1 << endl;
        return 0;
      }
      cout << -1 << endl;
      return 0;
    }
    ll p = bsgs(A[0] + A[3], B[k] * modinv(A[k]));
    if (matpow(A, p + 1) == B) {
      mat R = A;
      // 嘘解法
      for (int i = 1; i <= min(p, 100LL); i++) {
        if (R == B) {
          cout << i << endl;
          return 0;
        }
        R = mul(R, A);
      }
      cout << p + 1 << endl;
    } else {
      cout << -1 << endl;
    }
    return 0;
  }
  ll p = bsgs_except_zero(det(A), det(B));
  if (p == -1) {
    cout << -1 << endl;
    return 0;
  }
  ll T = bsgs_except_zero(det(A), 1);
  // (A^T)^k = B^A^{-N%T}
  ll k = bsgs_mat(matpow(A, T), mul(B, matpow(inv(A), p)));
  if (k == -1) {
    cout << -1 << endl;
    return 0;
  }
  cout << p + T*k << endl;
}