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 nmod = 1000000007;
void imprimirVector(vector<long long> v) {
  if (!v.empty()) {
    int p = v.size();
    cout << "[";
    for (long long i = 0; i < (long long)(p - 1); i++) cout << v[i] << ",";
    cout << v[p - 1] << "]" << endl;
  } else
    cout << "[]" << endl;
}
long long toNumber(string s) {
  long long Number;
  if (!(istringstream(s) >> Number)) Number = 0;
  return Number;
}
string toString(long long number) {
  ostringstream ostr;
  ostr << number;
  return ostr.str();
}
long long potLogMod(long long x, long long y) {
  long long ans = 1;
  while (y > 0) {
    if (y % 2) ans = (x * ans) % nmod;
    x = (x * x) % nmod;
    y /= 2;
  }
  return ans;
}
long long invMod(long long a) { return potLogMod(a, nmod - 2); }
long long mod(long long a) { return ((a % nmod) + nmod) % nmod; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long a, b, n, x;
  while (cin >> a >> b >> n >> x) {
    if (a == 1)
      cout << mod(x + mod(mod(b) * mod(n))) << "\n";
    else
      cout << mod(mod(potLogMod(a, n) * x) +
                  mod(mod(b * (potLogMod(a, n) - 1)) * invMod(a - 1)))
           << "\n";
  }
  return 0;
}
