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 mul(long long x, long long y) {
  return ((x % 1000000007) * (y % 1000000007)) % 1000000007;
}
long long sum(long long x, long long y) {
  return (x % 1000000007 + y % 1000000007) % 1000000007;
}
long long fp(long long x, long long y) {
  if (y == 0) return 1;
  long long t = fp(x, y / 2) % 1000000007;
  if (y & 1) {
    return mul(x, mul(t, t));
  } else {
    return mul(t, t);
  }
}
long long divi(long long x, long long y) {
  if (x == 1) {
    return y % 1000000007;
  }
  long long inv = fp((x - 1), 1000000007 - 2) % 1000000007;
  long long rr = fp(x, y) % 1000000007;
  return mul((rr - 1 + 1000000007) % 1000000007, inv);
}
int main() {
  long long A, B, x, n;
  cin >> A >> B >> n >> x;
  long long ans = mul(fp(A, n), fp(x, 1)) % 1000000007;
  ans = sum(ans, mul(divi(A, n), B));
  cout << ans % 1000000007;
  return 0;
}
