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 = 1000000007;
long long A, B, n, x;
long long mult_mod(long long a, long long b, long long p) {
  if (p <= 1e9)
    return a * b % p;
  else if (p <= (long long)1e12)
    return (((a * (b >> 20) % p) << 20) + (a * (b & ((1 << 20) - 1)))) % p;
  else {
    long long d = (long long)floor(a * (long double)b / p + 0.5);
    long long ret = (a * b - d * p) % p;
    if (ret < 0) ret += p;
    return ret;
  }
}
long long pow_mod(long long a, long long k, long long p) {
  long long r = 1;
  a %= p;
  for (; k; k >>= 1, a = mult_mod(a, a, p))
    if (k & 1) r = mult_mod(r, a, p);
  return r;
}
int main() {
  cin >> A >> B >> n >> x;
  long long ans;
  if (A != 1)
    ans =
        ((pow_mod(A, n, mod) * x) % mod +
         ((pow_mod(A, n, mod) - 1) * pow_mod(A - 1, mod - 2, mod)) % mod * B) %
        mod;
  else
    ans = (B * (n % mod) + x) % mod;
  ans = (ans + mod) % mod;
  cout << ans << '\n';
  return 0;
}
