#include <bits/stdc++.h>
using namespace std;
int64_t mx = 1e9 + 9;
int64_t po(int64_t a, int64_t b) {
  int64_t ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * (a % mx)) % mx;
    }
    a = a * a % mx;
    b = b >> 1;
  }
  return ans;
}
int main() {
  int64_t n, a, b, k, curr = 0;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  for (int64_t i = 0; i < k; i++) {
    if (s[i] == '+') {
      curr = (curr + (po(a, n - i) * (po(b, i) % mx)) % mx) % mx;
    } else {
      curr = (curr - (po(a, n - i) * po(b, i) % mx) + mx) % mx;
    }
  }
  int64_t qq = po(a, k);
  int64_t t = (po(qq, (mx - 2) % (mx)) % mx) * po(b, k) % mx;
  if (t == 1)
    curr = curr * ((n + 1) / k) % mx;
  else
    curr = (curr * po(t - 1, (mx - 2) % (mx)) % (mx)) *
           (po(t, (n + 1) / k) - 1) % mx;
  cout << curr << endl;
  return 0;
}

### Prompt

Your task is to create 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;
int64_t mx = 1e9 + 9;
int64_t po(int64_t a, int64_t b) {
  int64_t ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * (a % mx)) % mx;
    }
    a = a * a % mx;
    b = b >> 1;
  }
  return ans;
}
int main() {
  int64_t n, a, b, k, curr = 0;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  for (int64_t i = 0; i < k; i++) {
    if (s[i] == '+') {
      curr = (curr + (po(a, n - i) * (po(b, i) % mx)) % mx) % mx;
    } else {
      curr = (curr - (po(a, n - i) * po(b, i) % mx) + mx) % mx;
    }
  }
  int64_t qq = po(a, k);
  int64_t t = (po(qq, (mx - 2) % (mx)) % mx) * po(b, k) % mx;
  if (t == 1)
    curr = curr * ((n + 1) / k) % mx;
  else
    curr = (curr * po(t - 1, (mx - 2) % (mx)) % (mx)) *
           (po(t, (n + 1) / k) - 1) % mx;
  cout << curr << endl;
  return 0;
}

```