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>
#pragma warning(disable : 4996)
using namespace std;
const long long MOD = 1e9 + 7;
long long A, B, n, x;
void solve() {
  long long t[2][2] = {A, 0, B, 1};
  long long ans[2][2] = {x, 1, 0, 0};
  long long tmp[2][2] = {0};
  while (n) {
    if (n & 1) {
      for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++) tmp[i][j] = ans[i][j];
      memset(ans, 0, sizeof(ans));
      for (int i = 0; i < 2; ++i)
        for (int j = 0; j < 2; ++j)
          for (int k = 0; k < 2; ++k)
            ans[i][j] = (ans[i][j] + tmp[i][k] * t[k][j]) % MOD;
    }
    for (int i = 0; i < 2; ++i)
      for (int j = 0; j < 2; ++j) tmp[i][j] = t[i][j];
    memset(t, 0, sizeof(t));
    for (int i = 0; i < 2; ++i)
      for (int j = 0; j < 2; ++j)
        for (int k = 0; k < 2; ++k)
          t[i][j] = (t[i][j] + tmp[i][k] * tmp[k][j]) % MOD;
    n /= 2;
  }
  cout << ans[0][0] << endl;
}
int main() {
  cin >> A >> B >> n >> x;
  solve();
}
