#include <bits/stdc++.h>
using namespace std;
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long ans = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return ans;
}
int main() {
  long long a1, b1, a2, b2, l, r, xx, yy;
  cin >> a1 >> b1 >> a2 >> b2 >> l >> r;
  long long d = exgcd(a1, a2, xx, yy);
  if ((b2 - b1) % d) {
    cout << 0 << endl;
    return 0;
  }
  long long x = a2 / d;
  xx = (b2 - b1) / d * xx;
  xx = (xx % x + x) % x;
  l = max(l, max(b1, b2));
  long long ans = 0, cnt = a1 * xx + b1, lcm = a1 / d * a2;
  if (l > r) {
    cout << 0 << endl;
    return 0;
  }
  if (cnt <= r) ans += (r - cnt) / lcm + 1;
  if (cnt < l) ans -= (l - cnt - 1) / lcm + 1;
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

You are given two arithmetic progressions: a1k + b1 and a2l + b2. Find the number of integers x such that L ≤ x ≤ R and x = a1k' + b1 = a2l' + b2, for some integers k', l' ≥ 0.

Input

The only line contains six integers a1, b1, a2, b2, L, R (0 < a1, a2 ≤ 2·109, - 2·109 ≤ b1, b2, L, R ≤ 2·109, L ≤ R).

Output

Print the desired number of integers x.

Examples

Input

2 0 3 3 5 21


Output

3


Input

2 4 3 0 6 17


Output

2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long ans = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return ans;
}
int main() {
  long long a1, b1, a2, b2, l, r, xx, yy;
  cin >> a1 >> b1 >> a2 >> b2 >> l >> r;
  long long d = exgcd(a1, a2, xx, yy);
  if ((b2 - b1) % d) {
    cout << 0 << endl;
    return 0;
  }
  long long x = a2 / d;
  xx = (b2 - b1) / d * xx;
  xx = (xx % x + x) % x;
  l = max(l, max(b1, b2));
  long long ans = 0, cnt = a1 * xx + b1, lcm = a1 / d * a2;
  if (l > r) {
    cout << 0 << endl;
    return 0;
  }
  if (cnt <= r) ans += (r - cnt) / lcm + 1;
  if (cnt < l) ans -= (l - cnt - 1) / lcm + 1;
  cout << ans << endl;
  return 0;
}

```