#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T, typename TT>
inline ostream &operator<<(ostream &os, const pair<T, TT> &t) {
  return os << t.first << " " << t.second;
}
template <typename T>
inline ostream &operator<<(ostream &os, const vector<T> &t) {
  for (auto i : t) os << i << " ";
  return os;
}
template <typename T>
inline istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T1, typename T2>
inline istream &operator>>(istream &is, pair<T1, T2> &t) {
  is >> t.first >> t.second;
  return is;
}
const long long mod = 1e9 + 9;
long long mpow(long long n, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) (ans *= n) %= mod;
    (n *= n) %= mod, p /= 2;
  }
  return ans;
}
long long inv(long long n) { return mpow(n % mod, mod - 2); }
void solve() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long q = (n + 1) / k;
  string s;
  cin >> s;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    (ans += mod * mod + (s[i] == '+' ? 1 : -1) * mpow(a, n - i) * mpow(b, i)) %=
        mod;
  };
  long long z = mpow(b, k) * inv(mpow(a, k)) % mod;
  if (z == 1) {
    cout << (ans * q) % mod << '\n';
    return;
  }
  (ans *= inv(z - 1 + mod)) %= mod;
  (ans *= mpow(z, q) - 1 + mod) %= mod;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

You are given two integers a and b. Moreover, you are given a sequence s_0, s_1, ..., s_{n}. All values in s are integers 1 or -1. It's known that sequence is k-periodic and k divides n+1. In other words, for each k ≤ i ≤ n it's satisfied that s_{i} = s_{i - k}.

Find out the non-negative remainder of division of ∑ _{i=0}^{n} s_{i} a^{n - i} b^{i} by 10^{9} + 9.

Note that the modulo is unusual!

Input

The first line contains four integers n, a, b and k (1 ≤ n ≤ 10^{9}, 1 ≤ a, b ≤ 10^{9}, 1 ≤ k ≤ 10^{5}).

The second line contains a sequence of length k consisting of characters '+' and '-'. 

If the i-th character (0-indexed) is '+', then s_{i} = 1, otherwise s_{i} = -1.

Note that only the first k members of the sequence are given, the rest can be obtained using the periodicity property.

Output

Output a single integer — value of given expression modulo 10^{9} + 9.

Examples

Input

2 2 3 3
+-+


Output

7


Input

4 1 5 1
-


Output

999999228

Note

In the first example:

(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = 2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2} = 7

In the second example:

(∑ _{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 ≡ 999999228 \pmod{10^{9} + 9}.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T, typename TT>
inline ostream &operator<<(ostream &os, const pair<T, TT> &t) {
  return os << t.first << " " << t.second;
}
template <typename T>
inline ostream &operator<<(ostream &os, const vector<T> &t) {
  for (auto i : t) os << i << " ";
  return os;
}
template <typename T>
inline istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T1, typename T2>
inline istream &operator>>(istream &is, pair<T1, T2> &t) {
  is >> t.first >> t.second;
  return is;
}
const long long mod = 1e9 + 9;
long long mpow(long long n, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) (ans *= n) %= mod;
    (n *= n) %= mod, p /= 2;
  }
  return ans;
}
long long inv(long long n) { return mpow(n % mod, mod - 2); }
void solve() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long q = (n + 1) / k;
  string s;
  cin >> s;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    (ans += mod * mod + (s[i] == '+' ? 1 : -1) * mpow(a, n - i) * mpow(b, i)) %=
        mod;
  };
  long long z = mpow(b, k) * inv(mpow(a, k)) % mod;
  if (z == 1) {
    cout << (ans * q) % mod << '\n';
    return;
  }
  (ans *= inv(z - 1 + mod)) %= mod;
  (ans *= mpow(z, q) - 1 + mod) %= mod;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```