#include <bits/stdc++.h>
using namespace std;
long long a, b, n, k, mod = 1000 * 1000 * 1000 + 9;
string s;
long long get_number(char c) { return (c == '+') ? 1 : (mod - 1); }
long long bin_pow(long long a, long long n) {
  if (!n) return 1;
  long long res = bin_pow(a, n >> 1);
  res = (res * res) % mod;
  if (n & 1) res = (res * a) % mod;
  return res;
}
long long ab;
int main() {
  cin >> n >> a >> b >> k >> s;
  ab = (b * bin_pow(a, mod - 2)) % mod;
  long long pw = bin_pow(a, n), sum = 0;
  for (int i = 0; i < k; ++i) {
    sum = (sum + get_number(s[i]) * pw) % mod;
    pw = (pw * ab) % mod;
  }
  ab = bin_pow(ab, k);
  long long s = n / k;
  if (ab == 1) {
    cout << (sum * (s + 1)) % mod;
    return 0;
  }
  long long res = bin_pow(ab, s + 1);
  res = (res + mod - 1) % mod;
  res = (res * bin_pow(ab - 1, mod - 2)) % mod;
  cout << (res * sum) % mod;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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>
using namespace std;
long long a, b, n, k, mod = 1000 * 1000 * 1000 + 9;
string s;
long long get_number(char c) { return (c == '+') ? 1 : (mod - 1); }
long long bin_pow(long long a, long long n) {
  if (!n) return 1;
  long long res = bin_pow(a, n >> 1);
  res = (res * res) % mod;
  if (n & 1) res = (res * a) % mod;
  return res;
}
long long ab;
int main() {
  cin >> n >> a >> b >> k >> s;
  ab = (b * bin_pow(a, mod - 2)) % mod;
  long long pw = bin_pow(a, n), sum = 0;
  for (int i = 0; i < k; ++i) {
    sum = (sum + get_number(s[i]) * pw) % mod;
    pw = (pw * ab) % mod;
  }
  ab = bin_pow(ab, k);
  long long s = n / k;
  if (ab == 1) {
    cout << (sum * (s + 1)) % mod;
    return 0;
  }
  long long res = bin_pow(ab, s + 1);
  res = (res + mod - 1) % mod;
  res = (res * bin_pow(ab - 1, mod - 2)) % mod;
  cout << (res * sum) % mod;
  return 0;
}

```