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
#include <bits/stdc++.h>
using namespace std;
const unsigned long long MOD = 1e9 + 7;
struct Matrix {
  unsigned long long a[2][2];
  Matrix() { memset(a, 0, sizeof a); }
  Matrix operator+(Matrix o) {
    Matrix ans;
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++) ans.a[i][j] = a[i][j] + o.a[i][j];
    ans.modulate();
    return ans;
  }
  Matrix operator*(Matrix o) {
    Matrix ans;
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++) ans.a[i][j] += a[i][k] * o.a[k][j];
    ans.modulate();
    return ans;
  }
  void modulate() {
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++) a[i][j] %= MOD;
  }
  void eye() {
    memset(a, 0, sizeof a);
    for (int i = 0; i < 2; i++) a[i][i] = 1;
  }
};
Matrix mod_pow(Matrix a, unsigned long long e) {
  Matrix ans;
  ans.eye();
  while (e) {
    if (e & 1) {
      ans = ans * a;
    }
    a = a * a;
    e >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  unsigned long long A, B, n, x;
  cin >> A >> B >> n >> x;
  Matrix ans;
  ans.a[0][0] = A;
  ans.a[0][1] = B;
  ans.a[1][1] = 1;
  ans = mod_pow(ans, n);
  cout << (ans.a[0][0] * x + ans.a[0][1]) % MOD << '\n';
  return 0;
}
