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>
template <typename T>
void pv(const T& v, const std::string c = "v:", std::ostream& s = std::cout) {
  s << c << "\n";
  for (const auto x : v) {
    s << x << " ";
  }
  s << "\n";
}
const unsigned long long int P = 1000000007ULL;
unsigned long long int power(unsigned long long int A,
                             unsigned long long int n) {
  if (n == 0) {
    return 1ULL;
  }
  if (n % 2 == 1) {
    return (A * power(A, n - 1)) % P;
  }
  unsigned long long int B = power(A, n / 2);
  return (B * B) % P;
}
int main() {
  unsigned long long int A, B, n, x;
  std::cin >> A >> B >> n >> x;
  if (n == 1ULL) {
    std::cout << ((A * x) % P + B) % P << "\n";
    return 0;
  }
  unsigned long long int Apn = power(A, n);
  unsigned long long int res = (Apn * x) % P;
  unsigned long long int res3;
  if (A != 1ULL) {
    unsigned long long int res2 = (Apn + P - 1ULL) % P;
    unsigned long long int obr = power(A - 1, P - 2);
    res3 = (res2 * obr) % P;
  } else {
    res3 = (A * n) % P;
  }
  unsigned long long int res4 = (res3 * B) % P;
  std::cout << (res4 + res) % P << "\n";
  return 0;
}
