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;
double gcd(double a, double b) { return a < 0.01 ? b : gcd(fmod(b, a), a); }
long long bpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
vector<int> parent, rnk;
void make_set(int v) {
  parent[v] = v;
  rnk[v] = 0;
}
int find_set(int v) {
  if (v == parent[v]) return v;
  return parent[v] = find_set(parent[v]);
}
void union_sets(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (rnk[a] < rnk[b]) swap(a, b);
    parent[b] = a;
    if (rnk[a] == rnk[b]) rnk[a]++;
  }
}
struct edge {
  int u, v, w;
  bool operator<(edge const& other) { return w < other.w; }
};
long long int n, k, y, h, t;
long long int kv, kv1, q, r, x, z, w, x1, x2, t1, t2, p;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> p >> q >> n >> x;
  if (p != 1)
    cout << ((bpow(p, n, (long long int)1000000007) * x) %
                 (long long int)1000000007 +
             (bpow(p, n, (long long int)1000000007) +
              (long long int)1000000007 - 1) *
                 q % (long long int)1000000007 *
                 bpow(p - 1, (long long int)1000000007 - 2,
                      (long long int)1000000007) %
                 (long long int)1000000007) %
                (long long int)1000000007;
  else
    cout << (x % (long long int)1000000007 +
             n % (long long int)1000000007 * q % (long long int)1000000007) %
                (long long int)1000000007;
  return 0;
}
