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;
long long int powermod(long long int B, long long int P, long long int M) {
  long long int r = 1;
  if (B >= M) B %= M;
  while (P > 0) {
    if ((P & 1) == 1) {
      r *= B;
      if (r >= M) r %= M;
    }
    P >>= 1;
    B *= B;
    if (B >= M) B %= M;
  }
  return r;
}
long long int modinv(long long int B, long long int M) {
  return powermod(B, M - 2, M);
}
int main() {
  long long int MOD = 1000000007LL;
  long long int A, B, N, x, res = 1;
  cin >> A >> B >> N >> x;
  if (A == 1) {
    res = x + B * (N % MOD);
  } else {
    long long int An = powermod(A, N, MOD);
    res = An * x;
    long long int tmp = B * (An - 1) % MOD;
    tmp = tmp * modinv(A - 1, MOD) % MOD;
    res += tmp;
  }
  res %= MOD;
  cout << res << endl;
  return 0;
}
