#include <bits/stdc++.h>
using namespace std;
long long int pw(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2) res = (res * x) % 1000000009;
    n /= 2;
    x = (x * x) % 1000000009;
  }
  return res;
}
int main() {
  long long int n, a, b, k;
  string st;
  cin >> n >> a >> b >> k >> st;
  long long int res = 0;
  for (int i = 0; i < k; i++) {
    if (st[i] == '-')
      res = (res - pw(a, n - i) * pw(b, i)) % 1000000009;
    else
      res = (res + pw(a, n - i) * pw(b, i)) % 1000000009;
  }
  if (res < 0) res += 1000000009;
  long long int q = (b * pw(a, 1000000009 - 2)) % 1000000009;
  q = pw(q, k);
  n = (n + 1) / k;
  long long int final = 0;
  if (q == 1)
    final = (res * n) % 1000000009;
  else {
    long long int x = (pw(q, n) - 1);
    long long int y = (q - 1);
    final = (x * pw(y, 1000000009 - 2)) % 1000000009;
    final = (final * res) % 1000000009;
  }
  if (final < 0) final += 1000000009;
  cout << final;
}

### 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>
using namespace std;
long long int pw(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2) res = (res * x) % 1000000009;
    n /= 2;
    x = (x * x) % 1000000009;
  }
  return res;
}
int main() {
  long long int n, a, b, k;
  string st;
  cin >> n >> a >> b >> k >> st;
  long long int res = 0;
  for (int i = 0; i < k; i++) {
    if (st[i] == '-')
      res = (res - pw(a, n - i) * pw(b, i)) % 1000000009;
    else
      res = (res + pw(a, n - i) * pw(b, i)) % 1000000009;
  }
  if (res < 0) res += 1000000009;
  long long int q = (b * pw(a, 1000000009 - 2)) % 1000000009;
  q = pw(q, k);
  n = (n + 1) / k;
  long long int final = 0;
  if (q == 1)
    final = (res * n) % 1000000009;
  else {
    long long int x = (pw(q, n) - 1);
    long long int y = (q - 1);
    final = (x * pw(y, 1000000009 - 2)) % 1000000009;
    final = (final * res) % 1000000009;
  }
  if (final < 0) final += 1000000009;
  cout << final;
}

```