#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e+9 + 7;
long long f(long long a, long long b, long long x) {
  return (a * x % mod + b) % mod;
}
long long p(long long x, long long n, long long a, long long b) {
  if (n == 0) return x % mod;
  long long res = p(x, n / 2, a * a % mod, (a * b % mod + b) % mod);
  if (n % 2 == 1) res = f(a, b, res) % mod;
  return res;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  cout << p(x, n, a, b);
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e+9 + 7;
long long f(long long a, long long b, long long x) {
  return (a * x % mod + b) % mod;
}
long long p(long long x, long long n, long long a, long long b) {
  if (n == 0) return x % mod;
  long long res = p(x, n / 2, a * a % mod, (a * b % mod + b) % mod);
  if (n % 2 == 1) res = f(a, b, res) % mod;
  return res;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  cout << p(x, n, a, b);
  return 0;
}

```