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 = 1000000007;
long double PI = 4 * atan(1);
long long a, b, n, x, ans;
long long bigpow(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long r = bigpow(a, b / 2);
  long long ret = (r * r) % MOD;
  if (b % 2 == 0) {
    return ret;
  } else {
    return (ret * a) % MOD;
  }
}
long long inv(long long k) { return bigpow(k, MOD - 2); }
long long solve() {
  if (a == 1) return n % MOD;
  return ((bigpow(a, n) - 1) * inv(a - 1)) % MOD;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> a >> b >> n >> x;
  ans = ((bigpow(a, n) * x) % MOD + (b * (solve())) % MOD) % MOD;
  cout << ans;
  return 0;
}
