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 a, b, x, n;
void input() {
  cin >> a >> b >> n >> x;
  return;
}
long long powmod(long long a, long long b) {
  if (b == 1) return a;
  if (b < 1) return 1;
  long long t = powmod(a, b / 2) % 1000000007;
  t = (t * t) % 1000000007;
  if (b % 2 == 1) {
    t = (t * a) % 1000000007;
  }
  return t;
}
void solve() {
  if (a == 1) {
    cout << (b * (n % 1000000007) + x) % 1000000007;
    return;
  }
  long long tmp = powmod(a, n);
  long long ans = ((tmp - 1) * powmod(a - 1, 1000000007 - 2)) % 1000000007;
  tmp = (tmp * x) % 1000000007;
  ans = (ans * b) % 1000000007;
  cout << (tmp + ans) % 1000000007;
  return;
}
void output() { return; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  {
    input();
    solve();
    output();
  }
  return 0;
}
