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>
long long mod = 1000000007;
using namespace std;
long long fpow(long long a, long long b) {
  long long r = 1;
  while (b > 0) {
    if (b & 1) r *= a, r %= mod;
    a *= a;
    a %= mod;
    b >>= 1ll;
  }
  return r;
}
long long mulmod(long long a, long long b) {
  if (mod <= 1000000009) return a * b % mod;
  long long res = 0;
  while (a > 0) {
    if (a & 1) {
      res += b;
      if (res >= mod) res -= mod;
    }
    a >>= 1;
    b <<= 1;
    if (b >= mod) b -= mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long A, B, n, x;
  cin >> A >> B >> n >> x;
  long long res = fpow(A, n);
  res *= x;
  res %= mod;
  long long res2 = B % mod;
  res2 *= ((fpow(A, n) - 1 + mod) % mod);
  res2 %= mod;
  res2 *= (fpow(A - 1, mod - 2));
  res2 %= mod;
  res += res2;
  res %= mod;
  if (A == 1) {
    res = n;
    res %= mod;
    res *= B;
    res %= mod;
    res += x;
    res %= mod;
  }
  cout << res << endl;
  return 0;
}
