#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long n) {
  if (n == 0) {
    return 1;
  }
  long long ans = power(x, n / 2);
  ans = (ans * ans) % 1000000009;
  if (n % 2) {
    ans = (ans * x) % 1000000009;
  }
  return ans;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  int p[k];
  char c;
  for (int i = 0; i < k; i++) {
    cin >> c;
    if (c == '+') {
      p[i] = 1;
    } else {
      p[i] = -1;
    }
  }
  long long temp =
      (((power(b, n + 1) - power(a, n + 1) + 1000000009) % 1000000009) *
       (power((power(b, k) - power(a, k) + 1000000009) % 1000000009,
              1000000009 - 2))) %
      1000000009;
  if (power(b, k) == power(a, k)) {
    temp = (((n + 1) / k) * power(a, n + 1 - k)) % 1000000009;
  }
  long long x = (power(a, k - 1) * power(b, 0)) % 1000000009;
  long long ans = (p[0] * x + 1000000009) % 1000000009;
  for (int i = 1; i < k; i++) {
    x = (((x * (b)) % 1000000009) * power(a, 1000000009 - 2)) % 1000000009;
    ans = (ans + p[i] * x + 1000000009) % 1000000009;
  }
  cout << ((ans * temp) + 1000000009) % 1000000009 << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 power(long long x, long long n) {
  if (n == 0) {
    return 1;
  }
  long long ans = power(x, n / 2);
  ans = (ans * ans) % 1000000009;
  if (n % 2) {
    ans = (ans * x) % 1000000009;
  }
  return ans;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  int p[k];
  char c;
  for (int i = 0; i < k; i++) {
    cin >> c;
    if (c == '+') {
      p[i] = 1;
    } else {
      p[i] = -1;
    }
  }
  long long temp =
      (((power(b, n + 1) - power(a, n + 1) + 1000000009) % 1000000009) *
       (power((power(b, k) - power(a, k) + 1000000009) % 1000000009,
              1000000009 - 2))) %
      1000000009;
  if (power(b, k) == power(a, k)) {
    temp = (((n + 1) / k) * power(a, n + 1 - k)) % 1000000009;
  }
  long long x = (power(a, k - 1) * power(b, 0)) % 1000000009;
  long long ans = (p[0] * x + 1000000009) % 1000000009;
  for (int i = 1; i < k; i++) {
    x = (((x * (b)) % 1000000009) * power(a, 1000000009 - 2)) % 1000000009;
    ans = (ans + p[i] * x + 1000000009) % 1000000009;
  }
  cout << ((ans * temp) + 1000000009) % 1000000009 << endl;
}

```