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 mod = 1000000007;
long long res = 0;
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> > c(a.size(), vector<long long>(b[0].size()));
  for (int i = 0; i < a.size(); i++) {
    for (int k = 0; k < b.size(); k++) {
      if (a[i][k]) {
        for (int j = 0; j < b[0].size(); j++) {
          c[i][j] = (c[i][j] + a[i][k] * b[k][j] % mod) % mod;
        }
      }
    }
  }
  return c;
}
vector<vector<long long> > pow(vector<vector<long long> > a, long long n) {
  vector<vector<long long> > b(a.size(), vector<long long>(a.size()));
  for (int i = 0; i < a.size(); i++) b[i][i] = 1;
  while (n) {
    if (n & 1) b = mul(b, a);
    a = mul(a, a);
    n >>= 1;
  }
  return b;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &A, &B, &n, &x);
  vector<vector<long long> > a(2, vector<long long>(2));
  a[0][0] = A;
  a[0][1] = B;
  a[1][0] = 0;
  a[1][1] = 1;
  a = pow(a, n);
  long long res = (a[0][0] * x % mod + a[0][1]) % mod;
  printf("%I64d\n", res);
  return 0;
}
