#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long modexpo(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res % mod * a % mod) % mod;
    a = (a % mod * a % mod) % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long a, b, n, x, i;
  cin >> a >> b >> n >> x;
  if (n == 0) {
    cout << "0\n";
    return 0;
  }
  long long sum = b;
  long long temp = b;
  if (a == 1) {
    cout << (x % mod + (n % mod * b % mod) % mod) % mod;
  } else {
    temp = modexpo(a, n);
    sum = (temp % mod * x % mod) % mod;
    temp = ((temp % mod - 1 + mod) % mod * modexpo(a - 1, mod - 2) % mod * b %
            mod) %
           mod;
    sum = (sum % mod + (temp) % mod) % mod;
    cout << sum << "\n";
  }
}

### Prompt

Your challenge is to write 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;
long long modexpo(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res % mod * a % mod) % mod;
    a = (a % mod * a % mod) % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long a, b, n, x, i;
  cin >> a >> b >> n >> x;
  if (n == 0) {
    cout << "0\n";
    return 0;
  }
  long long sum = b;
  long long temp = b;
  if (a == 1) {
    cout << (x % mod + (n % mod * b % mod) % mod) % mod;
  } else {
    temp = modexpo(a, n);
    sum = (temp % mod * x % mod) % mod;
    temp = ((temp % mod - 1 + mod) % mod * modexpo(a - 1, mod - 2) % mod * b %
            mod) %
           mod;
    sum = (sum % mod + (temp) % mod) % mod;
    cout << sum << "\n";
  }
}

```