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;
void one(long long a[2][2]) {
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) a[i][j] = (i == j);
}
void mul(long long a[2][2], long long b[2][2], long long d) {
  long long res[2][2] = {{0}};
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++) {
        res[i][j] = (res[i][j] + ((a[i][k] % d) * (b[k][j] % d)) % d) % d;
      }
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) a[i][j] = res[i][j];
}
void pow(long long a[2][2], long long n, long long res[2][2], long long d) {
  one(res);
  while (n > 0) {
    if (n % 2 == 0) {
      mul(a, a, d);
      n /= 2;
    } else {
      mul(res, a, d);
      n--;
    }
  }
}
int main() {
  long long t, n, m, g, l, q, i, j, p, s = 0, ax, b, x, y, z, c;
  long long d = 1000000007;
  cin >> ax >> b >> n >> x;
  s = x;
  ax = ax % d;
  b = b % d;
  long long a[2][2];
  a[0][0] = ax;
  a[0][1] = b;
  a[1][0] = 0;
  a[1][1] = 1;
  long long ans[2][2];
  pow(a, n, ans, d);
  s = ((ans[0][0] * x) % d + ans[0][1]) % d;
  cout << s;
  return 0;
}
