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 INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int n = 2;
long long a, b, p, x;
vector<vector<long long> > operator*(const vector<vector<long long> > &a,
                                     const vector<vector<long long> > &b) {
  vector<vector<long long> > m(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++) {
        m[i][j] += (a[i][k] * b[k][j]) % MOD;
        m[i][j] %= MOD;
      }
  return m;
}
vector<vector<long long> > id() {
  vector<vector<long long> > m(n, vector<long long>(n));
  for (int i = 0; i < n; i++) m[i][i] = 1;
  return m;
}
long long expo(vector<vector<long long> > ele, long long e) {
  vector<vector<long long> > ans = id();
  while (e) {
    if (e & 1) ans = ans * ele;
    ele = ele * ele;
    e >>= 1;
  }
  long long resp = (ans[0][0] * x) % MOD;
  resp += (ans[1][0] * b) % MOD;
  resp %= MOD;
  return resp;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> a >> b >> p >> x;
  vector<vector<long long> > m(n, vector<long long>(n, 1LL));
  m[0][0] = a;
  m[0][1] = 0;
  cout << expo(m, p) << endl;
  return 0;
}
