#include <bits/stdc++.h>
using namespace std;
long long mod = (long long)1e9 + 7;
long long fast_expo(long long base, long long pw) {
  long long ans = 1;
  while (pw > 0) {
    if (pw % 2 == 1) ans = ans * base % mod;
    base = base * base % mod;
    pw = pw / 2;
  }
  return ans;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  long long inv1 = fast_expo(a - 1, mod - 2);
  long long an = fast_expo(a, n);
  long long bolu = (an - 1) * inv1 % mod;
  if (a == 1) bolu = n % mod;
  long long ilk = b * bolu % mod;
  long long iki = an * x % mod;
  long long res = ilk + iki;
  if (n == 1) {
    res = a * x % mod + b % mod;
  }
  cout << res % mod << endl;
}

### Prompt

Please formulate 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;
long long mod = (long long)1e9 + 7;
long long fast_expo(long long base, long long pw) {
  long long ans = 1;
  while (pw > 0) {
    if (pw % 2 == 1) ans = ans * base % mod;
    base = base * base % mod;
    pw = pw / 2;
  }
  return ans;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  long long inv1 = fast_expo(a - 1, mod - 2);
  long long an = fast_expo(a, n);
  long long bolu = (an - 1) * inv1 % mod;
  if (a == 1) bolu = n % mod;
  long long ilk = b * bolu % mod;
  long long iki = an * x % mod;
  long long res = ilk + iki;
  if (n == 1) {
    res = a * x % mod + b % mod;
  }
  cout << res % mod << endl;
}

```