[THE SxPLAY & KIVΛ - 漂流](https://soundcloud.com/kivawu/hyouryu)

[KIVΛ & Nikki Simmons - Perspectives](https://soundcloud.com/kivawu/perspectives)

With a new body, our idol Aroma White (or should we call her Kaori Minamiya?) begins to uncover her lost past through the OS space.

The space can be considered a 2D plane, with an infinite number of data nodes, indexed from 0, with their coordinates defined as follows:

  * The coordinates of the 0-th node is (x_0, y_0) 
  * For i > 0, the coordinates of i-th node is (a_x ⋅ x_{i-1} + b_x, a_y ⋅ y_{i-1} + b_y) 



Initially Aroma stands at the point (x_s, y_s). She can stay in OS space for at most t seconds, because after this time she has to warp back to the real world. She doesn't need to return to the entry point (x_s, y_s) to warp home.

While within the OS space, Aroma can do the following actions:

  * From the point (x, y), Aroma can move to one of the following points: (x-1, y), (x+1, y), (x, y-1) or (x, y+1). This action requires 1 second. 
  * If there is a data node at where Aroma is staying, she can collect it. We can assume this action costs 0 seconds. Of course, each data node can be collected at most once. 



Aroma wants to collect as many data as possible before warping back. Can you help her in calculating the maximum number of data nodes she could collect within t seconds?

Input

The first line contains integers x_0, y_0, a_x, a_y, b_x, b_y (1 ≤ x_0, y_0 ≤ 10^{16}, 2 ≤ a_x, a_y ≤ 100, 0 ≤ b_x, b_y ≤ 10^{16}), which define the coordinates of the data nodes.

The second line contains integers x_s, y_s, t (1 ≤ x_s, y_s, t ≤ 10^{16}) – the initial Aroma's coordinates and the amount of time available.

Output

Print a single integer — the maximum number of data nodes Aroma can collect within t seconds.

Examples

Input


1 1 2 3 1 0
2 4 20


Output


3

Input


1 1 2 3 1 0
15 27 26


Output


2

Input


1 1 2 3 1 0
2 2 1


Output


0

Note

In all three examples, the coordinates of the first 5 data nodes are (1, 1), (3, 3), (7, 9), (15, 27) and (31, 81) (remember that nodes are numbered from 0).

In the first example, the optimal route to collect 3 nodes is as follows: 

  * Go to the coordinates (3, 3) and collect the 1-st node. This takes |3 - 2| + |3 - 4| = 2 seconds. 
  * Go to the coordinates (1, 1) and collect the 0-th node. This takes |1 - 3| + |1 - 3| = 4 seconds. 
  * Go to the coordinates (7, 9) and collect the 2-nd node. This takes |7 - 1| + |9 - 1| = 14 seconds. 



In the second example, the optimal route to collect 2 nodes is as follows: 

  * Collect the 3-rd node. This requires no seconds. 
  * Go to the coordinates (7, 9) and collect the 2-th node. This takes |15 - 7| + |27 - 9| = 26 seconds. 



In the third example, Aroma can't collect any nodes. She should have taken proper rest instead of rushing into the OS space like that.
#include <bits/stdc++.h>
using namespace std;
const int base = 1000000000;
const int base_digits = 9;
struct bigint {
  vector<int> z;
  int sign;
  bigint() : sign(1) {}
  bigint(long long v) { *this = v; }
  bigint(const string& s) { read(s); }
  void operator=(const bigint& v) {
    sign = v.sign;
    z = v.z;
  }
  void operator=(long long v) {
    sign = 1;
    if (v < 0) sign = -1, v = -v;
    z.clear();
    for (; v > 0; v = v / base) z.push_back(v % base);
  }
  bigint operator+(const bigint& v) const {
    if (sign == v.sign) {
      bigint res = v;
      for (int i = 0, carry = 0; i < (int)max(z.size(), v.z.size()) || carry;
           ++i) {
        if (i == (int)res.z.size()) res.z.push_back(0);
        res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
        carry = res.z[i] >= base;
        if (carry) res.z[i] -= base;
      }
      return res;
    }
    return *this - (-v);
  }
  bigint operator-(const bigint& v) const {
    if (sign == v.sign) {
      if (abs() >= v.abs()) {
        bigint res = *this;
        for (int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
          res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
          carry = res.z[i] < 0;
          if (carry) res.z[i] += base;
        }
        res.trim();
        return res;
      }
      return -(v - *this);
    }
    return *this + (-v);
  }
  void operator*=(int v) {
    if (v < 0) sign = -sign, v = -v;
    for (int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
      if (i == (int)z.size()) z.push_back(0);
      long long cur = z[i] * (long long)v + carry;
      carry = (int)(cur / base);
      z[i] = (int)(cur % base);
    }
    trim();
  }
  bigint operator*(int v) const {
    bigint res = *this;
    res *= v;
    return res;
  }
  friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
    int norm = base / (b1.z.back() + 1);
    bigint a = a1.abs() * norm;
    bigint b = b1.abs() * norm;
    bigint q, r;
    q.z.resize(a.z.size());
    for (int i = a.z.size() - 1; i >= 0; i--) {
      r *= base;
      r += a.z[i];
      int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
      int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
      int d = ((long long)s1 * base + s2) / b.z.back();
      r -= b * d;
      while (r < 0) r += b, --d;
      q.z[i] = d;
    }
    q.sign = a1.sign * b1.sign;
    r.sign = a1.sign;
    q.trim();
    r.trim();
    return make_pair(q, r / norm);
  }
  friend bigint sqrt(const bigint& a1) {
    bigint a = a1;
    while (a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);
    int n = a.z.size();
    int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
    int norm = base / (firstDigit + 1);
    a *= norm;
    a *= norm;
    while (a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);
    bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
    firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
    int q = firstDigit;
    bigint res;
    for (int j = n / 2 - 1; j >= 0; j--) {
      for (;; --q) {
        bigint r1 =
            (r - (res * 2 * base + q) * q) * base * base +
            (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0);
        if (r1 >= 0) {
          r = r1;
          break;
        }
      }
      res *= base;
      res += q;
      if (j > 0) {
        int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
        int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
        int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
        q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
            (firstDigit * 2);
      }
    }
    res.trim();
    return res / norm;
  }
  bigint operator/(const bigint& v) const { return divmod(*this, v).first; }
  bigint operator%(const bigint& v) const { return divmod(*this, v).second; }
  void operator/=(int v) {
    if (v < 0) sign = -sign, v = -v;
    for (int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
      long long cur = z[i] + rem * (long long)base;
      z[i] = (int)(cur / v);
      rem = (int)(cur % v);
    }
    trim();
  }
  bigint operator/(int v) const {
    bigint res = *this;
    res /= v;
    return res;
  }
  int operator%(int v) const {
    if (v < 0) v = -v;
    int m = 0;
    for (int i = z.size() - 1; i >= 0; --i)
      m = (z[i] + m * (long long)base) % v;
    return m * sign;
  }
  void operator+=(const bigint& v) { *this = *this + v; }
  void operator-=(const bigint& v) { *this = *this - v; }
  void operator*=(const bigint& v) { *this = *this * v; }
  void operator/=(const bigint& v) { *this = *this / v; }
  bool operator<(const bigint& v) const {
    if (sign != v.sign) return sign < v.sign;
    if (z.size() != v.z.size()) return z.size() * sign < v.z.size() * v.sign;
    for (int i = z.size() - 1; i >= 0; i--)
      if (z[i] != v.z[i]) return z[i] * sign < v.z[i] * sign;
    return false;
  }
  bool operator>(const bigint& v) const { return v < *this; }
  bool operator<=(const bigint& v) const { return !(v < *this); }
  bool operator>=(const bigint& v) const { return !(*this < v); }
  bool operator==(const bigint& v) const {
    return !(*this < v) && !(v < *this);
  }
  bool operator!=(const bigint& v) const { return *this < v || v < *this; }
  void trim() {
    while (!z.empty() && z.back() == 0) z.pop_back();
    if (z.empty()) sign = 1;
  }
  bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }
  bigint operator-() const {
    bigint res = *this;
    res.sign = -sign;
    return res;
  }
  bigint abs() const {
    bigint res = *this;
    res.sign *= res.sign;
    return res;
  }
  long long longValue() const {
    long long res = 0;
    for (int i = z.size() - 1; i >= 0; i--) res = res * base + z[i];
    return res * sign;
  }
  friend bigint __gcd(const bigint& a, const bigint& b) {
    return b.isZero() ? a : __gcd(b, a % b);
  }
  friend bigint lcm(const bigint& a, const bigint& b) {
    return a / __gcd(a, b) * b;
  }
  void read(const string& s) {
    sign = 1;
    z.clear();
    int pos = 0;
    while (pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
      if (s[pos] == '-') sign = -sign;
      ++pos;
    }
    for (int i = s.size() - 1; i >= pos; i -= base_digits) {
      int x = 0;
      for (int j = max(pos, i - base_digits + 1); j <= i; j++)
        x = x * 10 + s[j] - '0';
      z.push_back(x);
    }
    trim();
  }
  friend istream& operator>>(istream& stream, bigint& v) {
    string s;
    stream >> s;
    v.read(s);
    return stream;
  }
  friend ostream& operator<<(ostream& stream, const bigint& v) {
    if (v.sign == -1) stream << '-';
    stream << (v.z.empty() ? 0 : v.z.back());
    for (int i = (int)v.z.size() - 2; i >= 0; --i)
      stream << setw(base_digits) << setfill('0') << v.z[i];
    return stream;
  }
  static vector<int> convert_base(const vector<int>& a, int old_digits,
                                  int new_digits) {
    vector<long long> p(max(old_digits, new_digits) + 1);
    p[0] = 1;
    for (int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * 10;
    vector<int> res;
    long long cur = 0;
    int cur_digits = 0;
    for (int i = 0; i < (int)a.size(); i++) {
      cur += a[i] * p[cur_digits];
      cur_digits += old_digits;
      while (cur_digits >= new_digits) {
        res.push_back(int(cur % p[new_digits]));
        cur /= p[new_digits];
        cur_digits -= new_digits;
      }
    }
    res.push_back((int)cur);
    while (!res.empty() && res.back() == 0) res.pop_back();
    return res;
  }
  static vector<long long> karatsubaMultiply(const vector<long long>& a,
                                             const vector<long long>& b) {
    int n = a.size();
    vector<long long> res(n + n);
    if (n <= 32) {
      for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) res[i + j] += a[i] * b[j];
      return res;
    }
    int k = n >> 1;
    vector<long long> a1(a.begin(), a.begin() + k);
    vector<long long> a2(a.begin() + k, a.end());
    vector<long long> b1(b.begin(), b.begin() + k);
    vector<long long> b2(b.begin() + k, b.end());
    vector<long long> a1b1 = karatsubaMultiply(a1, b1);
    vector<long long> a2b2 = karatsubaMultiply(a2, b2);
    for (int i = 0; i < k; i++) a2[i] += a1[i];
    for (int i = 0; i < k; i++) b2[i] += b1[i];
    vector<long long> r = karatsubaMultiply(a2, b2);
    for (int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
    for (int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];
    for (int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
    for (int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
    for (int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
    return res;
  }
  bigint operator*(const bigint& v) const {
    vector<int> a6 = convert_base(this->z, base_digits, 6);
    vector<int> b6 = convert_base(v.z, base_digits, 6);
    vector<long long> a(a6.begin(), a6.end());
    vector<long long> b(b6.begin(), b6.end());
    while (a.size() < b.size()) a.push_back(0);
    while (b.size() < a.size()) b.push_back(0);
    while (a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
    vector<long long> c = karatsubaMultiply(a, b);
    bigint res;
    res.sign = sign * v.sign;
    for (int i = 0, carry = 0; i < (int)c.size(); i++) {
      long long cur = c[i] + carry;
      res.z.push_back((int)(cur % 1000000));
      carry = (int)(cur / 1000000);
    }
    res.z = convert_base(res.z, 6, base_digits);
    res.trim();
    return res;
  }
};
bigint abs(bigint x) { return x < 0 ? -x : x; }
long long x[100], y[100], ax, ay, bx, by, xs, ys, t;
int c;
void solve() {
  int ans = 0;
  for (int i = 0; i <= c; i++) {
    long long tx = xs, ty = ys, l = 0;
    for (int j = i; j <= c; j++) {
      l += abs(x[j] - tx) + abs(y[j] - ty);
      tx = x[j], ty = y[j];
      if (l > t) break;
      ans = max(ans, j - i + 1);
      long long tx2 = tx, ty2 = ty, l2 = l;
      for (int k = i - 1; k >= 0; k--) {
        l2 += abs(x[k] - tx2) + abs(y[k] - ty2);
        tx2 = x[k], ty2 = y[k];
        if (l2 > t) break;
        ans = max(ans, j - i + 1 + i - k);
      }
    }
    tx = xs, ty = ys, l = 0;
    for (int j = i; j >= 0; j--) {
      l += abs(x[j] - tx) + abs(y[j] - ty);
      tx = x[j], ty = y[j];
      if (l > t) break;
      ans = max(ans, i - j + 1);
      long long tx2 = tx, ty2 = ty, l2 = l;
      for (int k = i + 1; k <= c; k++) {
        l2 += abs(x[k] - tx2) + abs(y[k] - ty2);
        tx2 = x[k], ty2 = y[k];
        if (l2 > t) break;
        ans = max(ans, i - j + 1 + k - i);
      }
    }
  }
  cout << ans << endl;
}
int main() {
  bigint x[100], y[100], ax, ay, bx, by, xs, ys, t;
  cin >> x[0] >> y[0] >> ax >> ay >> bx >> by >> xs >> ys >> t;
  while (c < 90) {
    x[c + 1] = x[c] * ax + bx;
    y[c + 1] = y[c] * ay + by;
    c++;
  }
  while (c >= 0 && abs(x[c] - xs) + abs(y[c] - ys) > t) c--;
  if (c == -1) {
    return puts("0"), 0;
  }
  for (int i = 0; i <= c; i++)
    ::x[i] = x[i].longValue(), ::y[i] = y[i].longValue();
  ::ax = ax.longValue(), ::ay = ay.longValue(), ::bx = bx.longValue(),
  ::by = by.longValue(), ::xs = xs.longValue(), ::ys = ys.longValue(),
  ::t = t.longValue();
  solve();
}
