#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
int counting(long long a, long long b) {
  long long total = 1;
  while (b > 0) {
    if (b & 1) total = (total * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return total;
}
int main() {
  long long a, b, n, x, s;
  cin >> a >> b >> n >> x;
  if (a == 1) {
    s = (n % mod * b) % mod + x;
    cout << s % mod;
    return 0;
  }
  s = counting(a, n);
  long long s1 = counting(a - 1, mod - 2);
  long long s2 = (s - 1 + mod) % mod;
  s = s * x % mod;
  s = (s + s1 * s2 % mod * b % mod) % mod;
  cout << s % mod;
  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 = 1e9 + 7;
int counting(long long a, long long b) {
  long long total = 1;
  while (b > 0) {
    if (b & 1) total = (total * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return total;
}
int main() {
  long long a, b, n, x, s;
  cin >> a >> b >> n >> x;
  if (a == 1) {
    s = (n % mod * b) % mod + x;
    cout << s % mod;
    return 0;
  }
  s = counting(a, n);
  long long s1 = counting(a - 1, mod - 2);
  long long s2 = (s - 1 + mod) % mod;
  s = s * x % mod;
  s = (s + s1 * s2 % mod * b % mod) % mod;
  cout << s % mod;
  return 0;
}

```