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;
int long long power(int long long x, int long long y, int long long p) {
  int long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
pair<int long long, int long long> gcdext(int long long a, int long long b) {
  if (b == 0) {
    pair<int long long, int long long> ans;
    ans.first = 1;
    ans.second = 0;
    return ans;
  }
  pair<int long long, int long long> temp = gcdext(b, a % b);
  pair<int long long, int long long> ans;
  ans.second = temp.first - (a / b) * temp.second;
  ans.first = temp.second;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int long long n, t, q, k, st, maxi, mini,
      cnt = 0, a, b, c, d, m, len, x, y, z, w, ans = 0, l, r, odd, even, p, no,
      end;
  string str, s;
  bool flag = false, flag2 = false;
  t = 1;
  while (t--) {
    cin >> a >> b >> n >> x;
    k = 1000000007;
    if (n == 1) {
      cout << ((a * x) % k + b) % k << endl;
      continue;
    }
    if (a == 1) {
      cout << (x % k + ((b % k) * (n % k)) % k) % k << endl;
      continue;
    }
    ans = (ans + power(a, n % (k - 1), k) * x) % k;
    pair<int long long, int long long> res = gcdext(a - 1, k);
    res.first = (res.first + k) % k;
    ans =
        (ans +
         (((res.first * b) % k) * (power(a, n % (k - 1), k) - 1 + k) % k) % k) %
        k;
    cout << ans << endl;
  }
}
