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>
const long long mod = 1e9 + 7;
const int maxn = 1e6 + 5;
const double eps = 1e-6;
using namespace std;
bool eq(const double &a, const double &b) { return fabs(a - b) < eps; }
bool ls(const double &a, const double &b) { return a + eps < b; }
bool le(const double &a, const double &b) { return eq(a, b) || ls(a, b); }
long long kpow(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int main() {
  long long a, b, n, x;
  scanf("%lld%lld%lld%lld", &a, &b, &n, &x);
  long long ans = 0;
  if (a == 1) {
    ans = (x % mod + (n % mod * b % mod) % mod) % mod;
    printf("%lld\n", ans);
  } else {
    ans = kpow(a, n) * x;
    ans %= mod;
    long long t = kpow(a - 1, mod - 2) * b;
    t %= mod;
    long long tt = kpow(a, n) - 1;
    if (tt < 0) tt += mod;
    t *= tt;
    ans = (ans % mod + t % mod) % mod;
    printf("%lld\n", ans);
  }
  return 0;
}
