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 fp(long long b, long long e) {
  if (e == 0) return 1LL;
  if (e == 1LL) return b;
  long long ret = fp(b, e / 2LL);
  ret = (ret * ret) % 1000000007;
  if (e % 2LL) ret = (ret * b) % 1000000007;
  return ret;
}
long long mul(long long a, long long b) {
  a %= 1000000007;
  b %= 1000000007;
  long long ret = (a * b);
  if (ret >= 1000000007) ret %= 1000000007;
  return ret;
}
long long add(long long a, long long b) {
  a %= 1000000007;
  b %= 1000000007;
  long long ret = a + b;
  if (ret >= 1000000007) ret %= 1000000007;
  return ret;
}
int main() {
  long long A, B, n, x;
  cin >> A >> B >> n >> x;
  if (n == 0) {
    cout << x << endl;
    return 0;
  }
  if (A == 1) {
    long long ans = x;
    ans = add(ans, mul(n % 1000000007, B));
    cout << ans << endl;
    return 0;
  }
  long long ans = 0;
  ans = add(ans, mul(x, fp(A, n)));
  long long tans = (fp(A, n) - 1LL + 1LL * 1000000007) % 1000000007;
  tans = mul(B, tans);
  tans = mul(tans, fp(A - 1, 1000000007 - 2));
  ans = add(ans, tans);
  cout << ans << endl;
  return 0;
}
