Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B, n and x find the value of g(n)(x) modulo 109 + 7.

Input

The only line contains four integers A, B, n and x (1 ≤ A, B, x ≤ 109, 1 ≤ n ≤ 1018) — the parameters from the problem statement.

Note that the given value n can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Output

Print the only integer s — the value g(n)(x) modulo 109 + 7.

Examples

Input

3 4 1 1


Output

7


Input

3 4 2 1


Output

25


Input

3 4 3 1


Output

79
#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> prepearing_for_philosophy(long long n, long long a,
                                                     long long b) {
  if (!n) return make_pair(1ll, 0ll);
  pair<long long, long long> t = prepearing_for_philosophy(n / 2, a, b);
  pair<long long, long long> ans =
      make_pair((t.first * t.first) % (long long)(1e9 + 7),
                (t.first * t.second + t.second) % (long long)(1e9 + 7));
  if (n % 2)
    ans = make_pair((ans.first * a) % (long long)(1e9 + 7),
                    (ans.second * a + b) % (long long)(1e9 + 7));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(0);
  long long n, x, a, b;
  cin >> a >> b >> n >> x;
  pair<long long, long long> good_luck;
  good_luck = prepearing_for_philosophy(n, a, b);
  {
    std::cout << (x * good_luck.first + good_luck.second) % (long long)(1e9 + 7)
              << '\n';
  };
  return 0;
}
