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 N = 2;
const int MOD = (int)1e9 + 7;
long long A, B, n, x;
vector<vector<long long> > mul(vector<vector<long long> >& A,
                               vector<vector<long long> >& B) {
  vector<vector<long long> > ret(N, vector<long long>(N));
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      for (int k = 0; k < N; k++)
        ret[i][j] = (ret[i][j] + A[i][k] * B[k][j] % MOD) % MOD;
  return ret;
}
vector<vector<long long> > matPow(vector<vector<long long> > A, long long exp) {
  vector<vector<long long> > ret(N, vector<long long>(N));
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) ret[i][j] = i == j;
  while (exp) {
    if (exp & 1) ret = mul(ret, A);
    A = mul(A, A);
    exp >>= 1;
  }
  return ret;
}
int main() {
  cin >> A >> B >> n >> x;
  vector<vector<long long> > init = {{x, 1}, {0, 0}};
  vector<vector<long long> > trans = {{A, 0}, {B, 1}};
  trans = matPow(trans, n);
  cout << mul(init, trans)[0][0] << '\n';
}
