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 long long int modulo = 1e9 + 7;
vector<vector<long long int> > operator*(vector<vector<long long int> > m1,
                                         vector<vector<long long int> > m2) {
  vector<vector<long long int> > m(2, vector<long long int>(2, 0));
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++)
        m[i][j] = (m[i][j] + m1[i][k] * m2[k][j]) % modulo;
  return m;
}
vector<vector<long long int> > eleva(vector<vector<long long int> > m,
                                     long long int e) {
  if (e == 0) {
    vector<vector<long long int> > r(2, vector<long long int>(2, 0));
    for (int i = 0; i < 2; i++) r[i][i] = 1;
    return r;
  }
  if (e % 2 == 0) return eleva(m * m, e / 2);
  return m * eleva(m * m, e / 2);
}
vector<long long int> operator*(vector<vector<long long int> > m,
                                vector<long long int> v) {
  vector<long long int> r(2, 0);
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) r[i] = (r[i] + m[i][j] * v[j]) % modulo;
  return r;
}
int main() {
  long long int a, b, n, x;
  cin >> a >> b >> n >> x;
  vector<vector<long long int> > m(2, vector<long long int>(2, 0));
  m[0][0] = a;
  m[0][1] = b;
  m[1][1] = 1;
  vector<long long int> v(2, 1);
  v[0] = x;
  cout << (eleva(m, n) * v)[0] << endl;
}
