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;
long long sigma(long long n) { return (n * (n + 1) / 2); }
long long MOD(long long x) {
  if (x >= 0) return x;
  return (-x);
}
bool great(long long a, long long b) { return a > b; }
bool p_sm(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first) {
    return (a.second < b.second);
  }
  return (a.first < b.first);
}
long long max3(long long a, long long b, long long c) {
  return (max(a, max(b, c)));
}
long long max4(long long a, long long b, long long c, long long d) {
  return (max(max(a, b), max(c, d)));
}
long long min3(long long a, long long b, long long c) {
  return (min(a, min(b, c)));
}
long long min4(long long a, long long b, long long c, long long d) {
  return (min(min(a, b), min(c, d)));
}
vector<long long> factors(long long x) {
  vector<long long> v;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      v.push_back(i);
      if (i != x / i) {
        v.push_back(x / i);
      }
    }
  }
  return v;
}
vector<long long> ispr;
vector<long long> min_prime;
unordered_map<long long, long long> prime_fact(long long x) {
  unordered_map<long long, long long> mp;
  while (x != 1) {
    mp[min_prime[x]]++;
    x /= min_prime[x];
  }
  return mp;
}
map<long long, long long> factorizee(long long n) {
  map<long long, long long> mp;
  for (long long i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      mp[i]++;
      n /= i;
    }
  }
  if (n != 1) {
    mp[n]++;
  }
  return mp;
}
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2) {
      ans = (ans * a) % m;
    }
    a = (a * a) % m;
    b = b / 2;
  }
  return ans;
}
bool f(long long n) {
  if ((n % 400) == 0) {
    return true;
  }
  if ((n % 100) && !(n % 4)) {
    return true;
  }
  return false;
}
int main() {
  long long a, b, n, x;
  cin >> a >> b >> n >> x;
  long long ans = 0;
  if (a == 1) {
    ans += ((n % 1000000007) * b);
    ans %= 1000000007;
  } else {
    ans += (power(a, n, 1000000007) - 1);
    ans *= power(a - 1, 1000000007 - 2, 1000000007);
    ans %= 1000000007;
    ans *= b;
    ans %= 1000000007;
  }
  ans += (power(a, n, 1000000007) * ((x) % 1000000007)) % 1000000007;
  cout << ans % 1000000007 << endl;
}
