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 MAXN = 2;
const int mod = 1e9 + 7;
struct mat {
  long long m[MAXN + 1][MAXN + 1];
};
mat matmul(mat A, mat B) {
  mat ans;
  for (int i = 1; i <= MAXN; i++) {
    for (int j = 1; j <= MAXN; j++) {
      ans.m[i][j] = 0;
      for (int k = 1; k <= MAXN; k++)
        ans.m[i][j] = (ans.m[i][j] + A.m[i][k] * B.m[k][j] % mod) % mod;
    }
  }
  return ans;
}
mat matqpow(mat A, long long k) {
  mat ans;
  for (int i = 1; i <= MAXN; i++) {
    for (int j = 1; j <= MAXN; j++) {
      if (i == j)
        ans.m[i][j] = 1;
      else
        ans.m[i][j] = 0;
    }
  }
  while (k) {
    if (k & 1) ans = matmul(ans, A);
    A = matmul(A, A);
    k >>= 1;
  }
  return ans;
}
int main() {
  long long A, B, n, x;
  ios::sync_with_stdio(false);
  cin >> A >> B >> n >> x;
  mat M;
  M.m[1][1] = A;
  M.m[1][2] = 0;
  M.m[2][1] = B;
  M.m[2][2] = 1;
  mat res = matqpow(M, n);
  long long ans = (res.m[1][1] * x % mod + res.m[2][1] % mod) % mod;
  cout << ans;
}
