#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long pow(long long aa, long long nn) {
  if (!nn) return 1;
  long long x = pow(aa, nn >> 1);
  x = x * x % mod;
  if (nn & 1) x = x * aa % mod;
  return x;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  long long p = pow(a, n);
  if (a == 1)
    cout << (x + n % mod * b % mod) % mod;
  else
    cout << (p * x % mod + (p - 1) * pow(a - 1, mod - 2) % mod * b % mod) % mod;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 1000000007;
long long pow(long long aa, long long nn) {
  if (!nn) return 1;
  long long x = pow(aa, nn >> 1);
  x = x * x % mod;
  if (nn & 1) x = x * aa % mod;
  return x;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  long long p = pow(a, n);
  if (a == 1)
    cout << (x + n % mod * b % mod) % mod;
  else
    cout << (p * x % mod + (p - 1) * pow(a - 1, mod - 2) % mod * b % mod) % mod;
  return 0;
}

```