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 = 1000000007;
long long modular_multi(long long a, long long b, long long c) {
  long long res, temp;
  res = 0, temp = a % c;
  while (b) {
    if (b & 1) {
      res += temp;
      if (res >= c) {
        res -= c;
      }
    }
    temp <<= 1;
    if (temp >= c) {
      temp -= c;
    }
    b >>= 1;
  }
  return res;
}
long long modular_exp(long long a, long long b, long long c) {
  long long res, temp;
  res = 1 % c, temp = a % c;
  while (b) {
    if (b & 1) {
      res = modular_multi(res, temp, c);
    }
    temp = modular_multi(temp, temp, c);
    b >>= 1;
  }
  return res;
}
long long quickpow(long long m, long long n, long long k) {
  long long b = 1;
  while (n > 0) {
    if (n & 1) b = (b * m) % k;
    n = n >> 1;
    m = (m * m) % k;
  }
  return b;
}
int main() {
  long long a, b, n, x;
  scanf("%lld%lld%lld%lld", &a, &b, &n, &x);
  long long ans = b % mod;
  long long shang = quickpow(a, n, mod);
  shang = (shang + mod - 1) % mod;
  ans = shang * ans % mod;
  long long xia = quickpow((a - 1), mod - 2, mod);
  ans = ans * xia % mod;
  long long ans1;
  ans1 = quickpow(a, n, mod);
  ans1 = ans1 * x % mod;
  ans = (ans + ans1) % mod;
  if (a == 1) {
    ans = modular_multi(b, n, mod);
    ans = (ans + x) % mod;
  }
  printf("%lld\n", ans);
}
