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;
unsigned long long A, B;
unsigned long long f(unsigned long long xx) {
  return ((A * xx) % MOD + B) % MOD;
}
unsigned long long solve(unsigned long long n, unsigned long long curr) {
  if (n == 0) return curr;
  return f(solve(n - 1, curr) % MOD) % MOD;
}
unsigned long long mod_pow(unsigned long long b, unsigned long long e) {
  unsigned long long ans = 1;
  while (e) {
    if (e & 1) ans = (ans * b) % MOD;
    b = (b * b) % MOD;
    e >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  unsigned long long x, n;
  cin >> A >> B >> n >> x;
  unsigned long long inter = mod_pow(A, n);
  unsigned long long aa = (inter * x) % MOD;
  unsigned long long bb = 0;
  if (A == 1)
    bb = n % MOD;
  else
    bb = ((inter - 1LL) % MOD * mod_pow(A - 1, MOD - 2)) % MOD;
  bb *= B;
  bb %= MOD;
  cout << (aa + bb) % MOD << '\n';
  return 0;
}
