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 int MOD = 1000000007;
struct f {
  long long A, B;
};
f operator*(const f& f1, const f& f2) {
  f ret;
  ret.A = (f1.A * f2.A) % MOD;
  ret.B = (f1.A * f2.B) % MOD;
  ret.B = (ret.B + f1.B) % MOD;
  return ret;
}
f ex(f x, long long n) {
  f y({1ll, 0ll});
  while (n) {
    if (n & 1) {
      y = (y * x);
    }
    n >>= 1;
    x = (x * x);
  }
  return y;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  f g1({a, b});
  f gn = ex(g1, n);
  long long ans = (gn.A * x) % MOD;
  ans = (ans + gn.B) % MOD;
  cout << ans << endl;
  return 0;
};
