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 double PI = acos(-1.0);
const long long MOD = 1000000007;
vector<vector<long long>> operator*(vector<vector<long long>> a,
                                    vector<vector<long long>> b) {
  int lin = a.size();
  int col = b[0].size();
  int dia = b.size();
  vector<vector<long long>> c(lin, vector<long long>(col));
  for (int i = 0; i < lin; i++) {
    for (int j = 0; j < col; j++) {
      for (int k = 0; k < dia; k++) {
        c[i][j] = (c[i][j] + (a[i][k] * b[k][j]) % MOD) % MOD;
      }
    }
  }
  return c;
}
vector<vector<long long>> id(long long n) {
  vector<vector<long long>> m(n, vector<long long>(n));
  for (int i = 0; i < n; i++) {
    m[i][i] = 1;
  }
  return m;
}
vector<vector<long long>> operator^(vector<vector<long long>> a, long long n) {
  if (n == 0) return id(a.size());
  vector<vector<long long>> c = a ^ (n / 2);
  c = c * c;
  if (n % 2) {
    c = a * c;
  }
  return c;
}
int main(void) {
  long long A, B, n, x;
  ios_base::sync_with_stdio(false);
  scanf("%lld %lld %lld %lld", &A, &B, &n, &x);
  vector<vector<long long>> base({{x}, {1}});
  vector<vector<long long>> transf({{A, B}, {0, 1}});
  printf("%lld\n", ((transf ^ n) * base)[0][0]);
  return 0;
}
