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 max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
long long a, b, n, x;
long long bigmod(long long a, long long p) {
  if (!p) return 1;
  long long t = bigmod(a, p >> 1);
  t = t * t % 1000000007;
  if (p & 1) t = t * a % 1000000007;
  return t;
}
long long rev(long long a) { return bigmod(a, 1000000007 - 2); }
int main() {
  cin >> a >> b >> n >> x;
  if (a == 1) {
    cout << (x + (n % 1000000007) * b) % 1000000007 << endl;
    return 0;
  }
  long long rt1 = x * bigmod(a, n) % 1000000007;
  long long rt2 = (b * (bigmod(a, n) + 1000000007 - 1) % 1000000007 *
                   rev(a - 1) % 1000000007);
  cout << (rt1 + rt2) % 1000000007 << endl;
  return 0;
}
