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>
const long mod = 1000000007;
using namespace std;
long long fast_pow(long long base, long long exp) {
  long long result = 1;
  while (exp > 0) {
    if (exp % 2 == 1) {
      result = (result * base) % mod;
    }
    base = (base * base) % mod;
    exp = exp / 2;
  }
  return result;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long modInverse(long long a, long long m) {
  long long x, y;
  gcdExtended(a, m, &x, &y);
  long long res = (x % m + m) % m;
  return res;
}
long long geom_serie(long long A, long long n) {
  if (A == 1) {
    return n % mod;
  }
  return ((fast_pow(A, n) - 1) * modInverse(A - 1, mod)) % mod;
}
int main() {
  long long A, B, n, x;
  cin >> A >> B >> n >> x;
  cout << ((fast_pow(A, n) * x) + (B * geom_serie(A, n))) % mod;
  return 0;
}
