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 long long mod = 1e9 + 7;
const long long N = 1e5 + 7;
string str;
char ch;
long long x, t, n, i, j, q, k, l, r, chk, idx, m, arr[N], a, b, ans;
const long long K = 2;
vector<vector<long long> > mul(vector<vector<long long> > A,
                               vector<vector<long long> > B) {
  vector<vector<long long> > C(K + 1, vector<long long>(K + 1));
  for (long long i = 1; i <= K; i++)
    for (long long j = 1; j <= K; j++)
      for (long long k = 1; k <= K; k++)
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;
  return C;
}
vector<vector<long long> > pow_(vector<vector<long long> > A, long long p) {
  if (p == 1) return A;
  if (p % 2) return mul(A, pow_(A, p - 1));
  vector<vector<long long> > X = pow_(A, p / 2);
  return mul(X, X);
}
long long power(long long N) {
  vector<long long> F_1(K + 1);
  F_1[1] = x;
  F_1[2] = b;
  vector<vector<long long> > T(K + 1, vector<long long>(K + 1));
  T[1][1] = a, T[1][2] = 1;
  T[2][1] = 0, T[2][2] = 1;
  T = pow_(T, N);
  long long res = 0;
  for (long long i = 1; i <= K; i++) res = (res + T[1][i] * F_1[i]) % mod;
  return res;
}
int main() {
  scanf("%lld", &a);
  scanf("%lld", &b);
  scanf("%lld", &n);
  scanf("%lld", &x);
  cout << power(n) << endl;
  return 0;
}
