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 MOD = 1e9 + 7;
long long int fexp(long long int b, long long int e) {
  if (e == 0) return 1;
  long long int x = fexp(b, e / 2);
  long long int quad_x = (x * x) % MOD;
  return (quad_x * (e % 2 ? b : 1)) % MOD;
}
long long int inv(long long a) { return fexp(a, MOD - 2); }
long long int add(long long int a, long long int b) {
  auto r = (a + b) % MOD;
  return r < 0 ? r + MOD : r;
}
long long int sumpg(long long int q, long long int n) {
  if (q == 1) {
    return (n) % MOD;
  }
  return ((add(fexp(q, n), -1)) * (inv(add(q, -1)))) % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int a, b, n, x;
  cin >> a >> b >> n >> x;
  long long int pt1 = (x * fexp(a, n)) % MOD;
  long long int pt2 = (b * sumpg(a, n)) % MOD;
  long long int ans = (pt1 + pt2) % MOD;
  cout << ans << '\n';
}
