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 = 1e9 + 7;
struct mat {
  long long m[2][2];
};
mat mul(mat a, mat b) {
  mat c;
  memset(c.m, 0, sizeof(c.m));
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++)
      for (int k = 0; k < 2; k++)
        c.m[i][j] = (c.m[i][j] + a.m[i][k] * b.m[k][j] % mod) % mod;
  return c;
}
mat qmod(mat a, long long n) {
  mat c;
  c.m[0][0] = c.m[1][1] = 1;
  c.m[0][1] = c.m[1][0] = 0;
  for (; n; n >>= 1) {
    if (n & 1) c = mul(c, a);
    a = mul(a, a);
  }
  return c;
}
int main() {
  long long n, a, b, x;
  cin >> a >> b >> n >> x;
  mat t;
  t.m[0][0] = a;
  t.m[0][1] = b;
  t.m[1][0] = 0;
  t.m[1][1] = 1;
  mat c = qmod(t, n);
  long long ans = c.m[0][0] * x % mod + c.m[0][1];
  cout << ans % mod;
  return 0;
}
