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 mod = 1e9 + 7;
long long power(long long x, long long y, long long p = mod) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long inv(long long x) { return power(x, mod - 2); }
int32_t main() {
  ios::sync_with_stdio(0);
  long long A, B, n, x;
  cin >> A >> B >> n >> x;
  long long dx = power(A, n, mod);
  if (A == 1)
    cout << add(x, mul(B, n)) << endl;
  else {
    cout << add(mul(dx, x), mul(B, mul(1 - dx, inv(1 - A)))) << endl;
  }
  return 0;
}
