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 A, B, n, x;
struct he {
  long long a[4][4];
} ans, base, c;
he mul(he a, he b) {
  for (int i = 1; i <= 2; i++)
    for (int j = 1; j <= 2; j++) c.a[i][j] = 0;
  for (int i = 1; i <= 2; i++)
    for (int j = 1; j <= 2; j++)
      for (int k = 1; k <= 2; k++)
        c.a[i][j] =
            (c.a[i][j] + a.a[i][k] * b.a[k][j] % 1000000007) % 1000000007;
  return c;
}
he get(long long t) {
  if (t == 1) return base;
  he tmp = get(t / 2);
  tmp = mul(tmp, tmp);
  if (t & 1) tmp = mul(tmp, base);
  return tmp;
}
int main() {
  scanf("%lld%lld%lld%lld", &A, &B, &n, &x);
  base.a[1][1] = A;
  base.a[1][2] = B;
  base.a[2][1] = 0;
  base.a[2][2] = 1;
  ans = get(n);
  printf("%lld\n",
         (ans.a[1][1] * x % 1000000007 + ans.a[1][2] +
          ans.a[2][1] * x % 1000000007 + ans.a[2][2] - 1 + 1000000007) %
             1000000007);
}
