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 gcdEx(long long a, long long b, long long &x, long long &y, long long &g) {
  if (!b) {
    x = 1;
    y = 0;
    g = a;
    return;
  } else {
    gcdEx(b, a % b, x, y, g);
    long long x1 = y;
    long long y1 = x - (a / b) * y;
    x = x1;
    y = y1;
  }
}
long long inverse(long long a, long long mod) {
  long long x, y, g;
  gcdEx(a, mod, x, y, g);
  return (((x % mod) + mod) % mod);
}
long long logExp(long long a, long long exp) {
  if (exp == 0) return 1;
  if (exp == 1) return a;
  if (exp % 2) return (a * logExp(a, exp - 1)) % 1000000007;
  long long expr = logExp(a, exp / 2);
  return (expr * expr) % 1000000007;
}
int main() {
  ios::sync_with_stdio(false);
  long long A, B, n, x;
  cin >> A >> B >> n >> x;
  long long expan = logExp(A, n);
  long long expx, pgsum;
  long long inv2 = inverse(2, 1000000007);
  if (A == 1) {
    expx = x;
    pgsum = ((((A + 1) * (n) % 1000000007) % 1000000007) * inv2) % 1000000007;
  } else {
    expx = (expan * x) % 1000000007;
    long long invden = inverse(A - 1, 1000000007);
    pgsum = ((expan - 1) * invden) % 1000000007;
  }
  cout << (expx + (pgsum * B) % 1000000007) % 1000000007 << endl;
}
