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 modn = 1e9 + 7;
const int maxn = 1e7 + 5;
long long pow_mod(long long a, long long n) {
  if (n == 0) return 1;
  long long x = pow_mod(a, n / 2);
  long long ans = x * x % modn;
  if (n & 1) ans = ans * a % modn;
  return ans;
}
void exgcd(long long a, long long b, long long &q, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    q = a;
    return;
  }
  exgcd(b, a % b, q, y, x);
  y -= a / b * x;
}
long long Inv(long long a) {
  long long q, x, y;
  exgcd(a, modn, q, x, y);
  if (q == 1) return (x % modn + modn) % modn;
}
int main() {
  long long a, b, n, x, cnt;
  while (cin >> a >> b >> n >> x) {
    long long ans1, ans2;
    ans1 = pow_mod(a, n) * x % modn;
    if (a != 1) {
      ans2 = (pow_mod(a, n) - 1) * Inv(a - 1) % modn;
      ans2 = ans2 * b % modn;
    } else {
      ans2 = n % modn;
      ans2 = ans2 * b % modn;
    }
    cout << ((ans1 + ans2) % modn + modn) % modn << endl;
  }
  return 0;
}
