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;
typedef long long ll;
int mod = 1e9 + 7;
ll a, b, n, x;
struct matrix {
  int n, m;
  ll a[2][2];
  matrix(int n = 2, int m = 2) : n(n), m(m) { memset(a, 0, sizeof(a)); }
  matrix mod_mul(matrix &b, ll mod) const {
    matrix tmp(n, b.m);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < b.m; j++) {
        for (int k = 0; k < m; k++) {
          tmp.a[i][j] = (tmp.a[i][j] + a[i][k] * b.a[k][j] % mod) % mod;
        }
      }
    }
    return tmp;
  }
  matrix mod_pow(ll nn, ll mod) const {
    matrix a = *this, tmp(n, n);
    for (int i = 0; i < n; i++) {
      tmp.a[i][i] = 1;
    }
    for (; nn > 0; nn >>= 1) {
      if (nn & 1) {
        tmp = tmp.mod_mul(a, mod);
      }
      a = a.mod_mul(a, mod);
    }
    return tmp;
  }
} mat, ans;
int main() {
  cin >> a >> b >> n >> x;
  mat.a[0][0] = a;
  mat.a[0][1] = b;
  mat.a[1][0] = 0;
  mat.a[1][1] = 1;
  ans = mat.mod_pow(n, mod);
  cout << (x * ans.a[0][0] + ans.a[0][1]) % mod << endl;
  return 0;
}
