#include <bits/stdc++.h>
using namespace std;
const long long P = 1E9 + 7;
long long quickpow(long long a, long long b, long long c) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % c;
    a = a * a % c;
    b >>= 1LL;
  }
  return res;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  if (n == 0) {
    cout << x << endl;
    return 0;
  }
  if (a == 0) {
    cout << b << endl;
    return 0;
  } else if (a == 1) {
    cout << (x % P + n % P * b % P) % P << endl;
  } else {
    long long ans = 0;
    ans += quickpow(a, n, P) * (x % P) % P;
    ans += ((b % P) * (quickpow(a, n, P) - 1 + P) % P) *
           quickpow(a - 1, P - 2, P) % P;
    ans %= P;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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 P = 1E9 + 7;
long long quickpow(long long a, long long b, long long c) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % c;
    a = a * a % c;
    b >>= 1LL;
  }
  return res;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  if (n == 0) {
    cout << x << endl;
    return 0;
  }
  if (a == 0) {
    cout << b << endl;
    return 0;
  } else if (a == 1) {
    cout << (x % P + n % P * b % P) % P << endl;
  } else {
    long long ans = 0;
    ans += quickpow(a, n, P) * (x % P) % P;
    ans += ((b % P) * (quickpow(a, n, P) - 1 + P) % P) *
           quickpow(a - 1, P - 2, P) % P;
    ans %= P;
    cout << ans << endl;
  }
  return 0;
}

```