#include <bits/stdc++.h>
using namespace std;
long long int pwr(long long int a, long long int n, long long int m) {
  long long int p = 1;
  while (n > 0) {
    if (n % 2 == 1) p = (p * a) % m;
    a = (a * a) % m;
    n = n / 2;
  }
  return p;
}
char str[1000000];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, a, b, k, s = 0, i;
  cin >> n >> a >> b >> k;
  long long int f =
      (((pwr(b, n + 1, (long long int)(1e9 + 9)) -
         pwr(a, n + 1, (long long int)(1e9 + 9)) + (long long int)(1e9 + 9)) %
        (long long int)(1e9 + 9)) *
       pwr(a, k, (long long int)(1e9 + 9))) %
      (long long int)(1e9 + 9);
  long long int g =
      (((pwr(b, k, (long long int)(1e9 + 9)) -
         pwr(a, k, (long long int)(1e9 + 9)) + (long long int)(1e9 + 9)) %
        (long long int)(1e9 + 9)) *
       pwr(a, n + 1, (long long int)(1e9 + 9))) %
      (long long int)(1e9 + 9);
  long long int r =
      ((f * pwr(g, (long long int)(1e9 + 9) - 2, (long long int)(1e9 + 9))) %
           (long long int)(1e9 + 9) +
       (long long int)(1e9 + 9)) %
      (long long int)(1e9 + 9);
  if (r == 0) r = (n + 1) / k;
  cin >> str;
  for (i = 0; i < k; i++) {
    if (str[i] == '+')
      s = (s +
           (pwr(a, n - i, (long long int)(1e9 + 9)) *
            pwr(b, i, (long long int)(1e9 + 9))) %
               (long long int)(1e9 + 9) +
           (long long int)(1e9 + 9)) %
          (long long int)(1e9 + 9);
    else
      s = (s -
           (pwr(a, n - i, (long long int)(1e9 + 9)) *
            pwr(b, i, (long long int)(1e9 + 9))) %
               (long long int)(1e9 + 9) +
           (long long int)(1e9 + 9)) %
          (long long int)(1e9 + 9);
  }
  cout << ((s * r) % (long long int)(1e9 + 9) + (long long int)(1e9 + 9)) %
              (long long int)(1e9 + 9)
       << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 int pwr(long long int a, long long int n, long long int m) {
  long long int p = 1;
  while (n > 0) {
    if (n % 2 == 1) p = (p * a) % m;
    a = (a * a) % m;
    n = n / 2;
  }
  return p;
}
char str[1000000];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, a, b, k, s = 0, i;
  cin >> n >> a >> b >> k;
  long long int f =
      (((pwr(b, n + 1, (long long int)(1e9 + 9)) -
         pwr(a, n + 1, (long long int)(1e9 + 9)) + (long long int)(1e9 + 9)) %
        (long long int)(1e9 + 9)) *
       pwr(a, k, (long long int)(1e9 + 9))) %
      (long long int)(1e9 + 9);
  long long int g =
      (((pwr(b, k, (long long int)(1e9 + 9)) -
         pwr(a, k, (long long int)(1e9 + 9)) + (long long int)(1e9 + 9)) %
        (long long int)(1e9 + 9)) *
       pwr(a, n + 1, (long long int)(1e9 + 9))) %
      (long long int)(1e9 + 9);
  long long int r =
      ((f * pwr(g, (long long int)(1e9 + 9) - 2, (long long int)(1e9 + 9))) %
           (long long int)(1e9 + 9) +
       (long long int)(1e9 + 9)) %
      (long long int)(1e9 + 9);
  if (r == 0) r = (n + 1) / k;
  cin >> str;
  for (i = 0; i < k; i++) {
    if (str[i] == '+')
      s = (s +
           (pwr(a, n - i, (long long int)(1e9 + 9)) *
            pwr(b, i, (long long int)(1e9 + 9))) %
               (long long int)(1e9 + 9) +
           (long long int)(1e9 + 9)) %
          (long long int)(1e9 + 9);
    else
      s = (s -
           (pwr(a, n - i, (long long int)(1e9 + 9)) *
            pwr(b, i, (long long int)(1e9 + 9))) %
               (long long int)(1e9 + 9) +
           (long long int)(1e9 + 9)) %
          (long long int)(1e9 + 9);
  }
  cout << ((s * r) % (long long int)(1e9 + 9) + (long long int)(1e9 + 9)) %
              (long long int)(1e9 + 9)
       << endl;
  return 0;
}

```