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 long long int mod = 1e9 + 7;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % mod;
    }
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t danger_close = clock();
  long long int a, b, n, x;
  cin >> a >> b >> n >> x;
  if (a == 1) {
    cout << (n % mod * b % mod + x % mod) % mod;
  } else {
    long long int elar = power(a, n);
    cout << ((elar * x) % mod +
             (((b * (elar - 1)) % mod) * power(a - 1, mod - 2)) % mod) %
                mod;
  }
  fprintf(stderr, "Total Time: %.3f\n",
          (double)(clock() - danger_close) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}
