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 double EPS = 1e-10;
const long long int MOD = 1000000007ll;
const long long int mod1 = 1000000009ll;
const long long int mod2 = 1100000009ll;
int INF = 2147483645;
long long int INFINF = 9223372036854775807;
int debug = 0;
using namespace std;
void print(int a[], int s, int e) {
  for (int i = s; i <= e; i++) cout << a[i] << " ";
  cout << "\n";
}
void print(vector<int> &v, int s, int e) {
  for (int i = s; i <= e; i++) cout << v[i] << " ";
  cout << "\n";
}
void print(vector<int> &v) {
  for (int x : v) cout << x << " ";
  cout << "\n";
}
long long int bit_count(long long int _x) {
  long long int _ret = 0;
  while (_x) {
    if (_x % 2 == 1) _ret++;
    _x /= 2;
  }
  return _ret;
}
long long int bit(long long int _mask, long long int _i) {
  return (_mask & (1 << _i)) == 0 ? 0 : 1;
}
long long int powermod(long long int _a, long long int _b, long long int _m) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
long long int add(long long int a, long long int b, long long int m = MOD) {
  long long int x = a + b;
  while (x >= m) x -= m;
  return x;
}
long long int sub(long long int a, long long int b, long long int m = MOD) {
  long long int x = a - b;
  while (x < 0) x += m;
  return x;
}
long long int mul(long long int a, long long int b, long long int m = MOD) {
  long long int x = a * b;
  x %= m;
  return x;
}
long long int A, B, n, x;
int main() {
  srand(time(NULL));
  cin >> A >> B >> n >> x;
  if (A == 1) {
    long long int ans = x;
    ans = add(x, mul(B, n % MOD));
    cout << ans;
    return 0;
  }
  long long int ans = 0;
  ans += mul(powermod(A, n, MOD), x);
  long long int temp = B;
  temp = mul(temp, sub(powermod(A, n, MOD), 1));
  temp = mul(temp, powermod(A - 1, MOD - 2, MOD));
  ans = add(temp, ans);
  cout << ans;
  return 0;
}
