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;
inline long long readLLD() {
  long long ret = 0;
  bool negg = false;
  char c;
  c = getchar();
  while ((c < '0' || c > '9') && c != '-') {
    c = getchar();
  }
  if (c == '-') {
    negg = true;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0');
    c = getchar();
  }
  if (negg) {
    ret = -ret;
  }
  return ret;
}
inline int readInt() {
  int ret = 0;
  bool negg = false;
  char c;
  c = getchar();
  while ((c < '0' || c > '9') && c != '-') {
    c = getchar();
  }
  if (c == '-') {
    negg = true;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0');
    c = getchar();
  }
  if (negg) {
    ret = -ret;
  }
  return ret;
}
long long a, b, n, q;
pair<long long, long long> cal(long long N) {
  if (N == 0) {
    return make_pair(1, 0);
  }
  pair<long long, long long> e = cal(N >> 1);
  pair<long long, long long> res = {
      (e.first * e.first) % 1000000007,
      (e.first * e.second + e.second) % 1000000007};
  if (N & 1) {
    res = {(res.first * a) % 1000000007, (res.second * a + b) % 1000000007};
  }
  return res;
}
int main() {
  cin >> a >> b >> n >> q;
  pair<long long, long long> we = cal(n);
  cout << (q * we.first + we.second) % 1000000007 << endl;
  return 0;
}
