#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000000 + 10;
const int mod = int(1e9) + 9;
long long quick_pow(long long base, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * base % mod;
    base = base * base % mod;
    n >>= 1;
  }
  return ans % mod;
}
char str[maxn];
int main() {
  int n, a, b, k;
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", str);
  while (k < int(1e4)) {
    for (int i = 0; i < k; i++) str[i + k] = str[i];
    k <<= 1;
  }
  long long ans = 0;
  int m = (n + 1) / k;
  long long st = n;
  long long ed = 0;
  if (m > 0) {
    long long tmp = 0;
    if (st - k > 0) {
      st = st - k;
      for (int i = 0; i < k; i++) {
        if (str[i] == '+')
          tmp = (tmp + quick_pow(a, k - i) * quick_pow(b, i) % mod + mod) % mod;
        else
          tmp = (tmp - quick_pow(a, k - i) * quick_pow(b, i) % mod + mod) % mod;
      }
      while (st > 0) {
        ans = (ans + (quick_pow(a, st) * quick_pow(b, ed) % mod) * tmp % mod +
               mod) %
              mod;
        st -= k;
        ed += k;
        if (st <= 0) {
          st += k;
          ed -= k;
          break;
        }
      }
    }
  }
  long long sum = 0;
  for (int i = st; i >= 0; i--) {
    if (str[st - i] == '+')
      sum = (sum + quick_pow(a, i) * quick_pow(b, n - i) % mod + mod) % mod;
    else
      sum = (sum - quick_pow(a, i) * quick_pow(b, n - i) % mod + mod) % mod;
  }
  ans = (ans + sum + mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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;
const int maxn = 1000000 + 10;
const int mod = int(1e9) + 9;
long long quick_pow(long long base, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * base % mod;
    base = base * base % mod;
    n >>= 1;
  }
  return ans % mod;
}
char str[maxn];
int main() {
  int n, a, b, k;
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", str);
  while (k < int(1e4)) {
    for (int i = 0; i < k; i++) str[i + k] = str[i];
    k <<= 1;
  }
  long long ans = 0;
  int m = (n + 1) / k;
  long long st = n;
  long long ed = 0;
  if (m > 0) {
    long long tmp = 0;
    if (st - k > 0) {
      st = st - k;
      for (int i = 0; i < k; i++) {
        if (str[i] == '+')
          tmp = (tmp + quick_pow(a, k - i) * quick_pow(b, i) % mod + mod) % mod;
        else
          tmp = (tmp - quick_pow(a, k - i) * quick_pow(b, i) % mod + mod) % mod;
      }
      while (st > 0) {
        ans = (ans + (quick_pow(a, st) * quick_pow(b, ed) % mod) * tmp % mod +
               mod) %
              mod;
        st -= k;
        ed += k;
        if (st <= 0) {
          st += k;
          ed -= k;
          break;
        }
      }
    }
  }
  long long sum = 0;
  for (int i = st; i >= 0; i--) {
    if (str[st - i] == '+')
      sum = (sum + quick_pow(a, i) * quick_pow(b, n - i) % mod + mod) % mod;
    else
      sum = (sum - quick_pow(a, i) * quick_pow(b, n - i) % mod + mod) % mod;
  }
  ans = (ans + sum + mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

```