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 maxn = 1e5 + 5;
const int maxlog = 25;
const long long mod = 1e9 + 7;
const int sq = 720;
const int inf = 1e9;
long long powmod(long long a, long long n) {
  if (n == 0) return 1;
  long long x = powmod(a, n / 2);
  long long cur = (x * x) % mod;
  if (n & 1) cur = (cur * a) % mod;
  return cur;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout.precision(20);
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  if (a == 1) {
    cout << (x + (n % mod) * b) % mod;
    return 0;
  }
  long long ans =
      ((((b * (powmod(a, n) - 1)) % mod) * powmod(a - 1, mod - 2)) % mod +
       (powmod(a, n) * x) % mod) %
      mod;
  cout << ans;
  return 0;
}
