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;
template <typename T>
void printv(vector<T> v) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
template <typename T>
void printA(T arr[], int n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << endl;
}
template <typename T>
void print(T var) {
  cout << var << endl;
}
const long long p = 1e9 + 7;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modinv(int a) { return power(a, p - 2); }
long long a, b, n, x;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> a >> b >> n >> x;
  long long num = power(a, n) - 1;
  if (num < 0) num += p;
  if (a != 1)
    num = (modinv(a - 1) * num) % p;
  else
    num = n % p;
  long long ans = (power(a, n) * (x % p)) % p + ((b % p) * num) % p;
  cout << ans % p << endl;
}
