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 int inf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3fll;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const int MAX = 1e6 + 10;
const int mod = 1e9 + 7;
vector<vector<long long> > mul(const vector<vector<long long> > &A,
                               const vector<vector<long long> > &B) {
  int n = ((int)(A).size()), nm = ((int)(B).size()), m = ((int)(B[0]).size());
  vector<vector<long long> > C(n, vector<long long>(m));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      for (int k = 0; k < nm; k++)
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;
  return C;
}
vector<vector<long long> > mpow(vector<vector<long long> > &A,
                                long long power) {
  int K = ((int)(A).size());
  vector<vector<long long> > iden(K, vector<long long>(K));
  for (int i = 0; i < K; i++) iden[i][i] = 1ll;
  while (power) {
    if (power & 1) iden = mul(iden, A);
    A = mul(A, A);
    power >>= 1;
  }
  return iden;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  vector<vector<long long> > m(1, vector<long long>(2));
  m[0][0] = x;
  m[0][1] = 1;
  vector<vector<long long> > m2(2, vector<long long>(2));
  m2[0][0] = a;
  m2[1][0] = b;
  m2[1][1] = 1;
  m2 = mpow(m2, n);
  m = mul(m, m2);
  cout << m[0][0] << endl;
  return 0;
}
