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;
vector<vector<long long> > mul(vector<vector<long long> > a,
                               vector<vector<long long> > b) {
  vector<vector<long long> > c(a.size(), vector<long long>(b[0].size(), 0));
  for (int i = 0; i < a.size(); i++) {
    for (int x = 0; x < b[0].size(); x++) {
      for (int j = 0; j < b.size(); j++)
        c[i][x] = (c[i][x] + a[i][j] * b[j][x]) % 1000000007;
    }
  }
  return c;
}
vector<vector<long long> > pow(vector<vector<long long> > a, long long x) {
  if (x == 1) return a;
  vector<vector<long long> > tmp = pow(a, x / 2);
  tmp = mul(tmp, tmp);
  if (x % 2) tmp = mul(tmp, a);
  return tmp;
}
int main() {
  long long a, b, n, x;
  scanf("%I64d%I64d%I64d%I64d", &a, &b, &n, &x);
  vector<vector<long long> > mat, vec;
  mat.push_back({a, 1});
  mat.push_back({0, 1});
  vec.push_back({x});
  vec.push_back({b});
  vec = mul(pow(mat, n), vec);
  printf("%I64d", vec[0][0]);
}
