#include <bits/stdc++.h>
using namespace std;
const long long mol = 1e9 + 9;
const int N = 100005;
char s[N];
long long n, a, ans, b, k;
long long w = 0;
long long ksm(long long a, long long k) {
  long long c = 1;
  while (k) {
    if (k & 1) c = (c * a) % mol;
    a = (a * a) % mol;
    k >>= 1;
  }
  return c;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", s);
  int o = (n + 1) / k;
  int A = ksm(a, mol - 2);
  for (int i = 0; i < k; i++) {
    if (s[i] == '-')
      w -= ksm(a, n - i) * ksm(b, i);
    else
      w += ksm(a, n - i) * ksm(b, i);
    w %= mol;
  }
  long long q = ksm(A, k) * ksm(b, k) % mol;
  if (q != 1) {
    ans = (w * (ksm(q, o) - 1) % mol * ksm(q - 1, mol - 2) % mol + mol) % mol;
  } else
    ans = w * o % mol;
  if ((n + 1) % k) {
    for (int i = o * k; i <= n; i++) {
      if (s[i] == '-')
        ans -= ksm(a, n - i) * ksm(b, i);
      else
        ans += ksm(a, n - i) * ksm(b, i);
      ans %= mol;
    }
  }
  printf("%I64d", (ans + mol) % mol);
  return 0;
}

### 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;
const long long mol = 1e9 + 9;
const int N = 100005;
char s[N];
long long n, a, ans, b, k;
long long w = 0;
long long ksm(long long a, long long k) {
  long long c = 1;
  while (k) {
    if (k & 1) c = (c * a) % mol;
    a = (a * a) % mol;
    k >>= 1;
  }
  return c;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", s);
  int o = (n + 1) / k;
  int A = ksm(a, mol - 2);
  for (int i = 0; i < k; i++) {
    if (s[i] == '-')
      w -= ksm(a, n - i) * ksm(b, i);
    else
      w += ksm(a, n - i) * ksm(b, i);
    w %= mol;
  }
  long long q = ksm(A, k) * ksm(b, k) % mol;
  if (q != 1) {
    ans = (w * (ksm(q, o) - 1) % mol * ksm(q - 1, mol - 2) % mol + mol) % mol;
  } else
    ans = w * o % mol;
  if ((n + 1) % k) {
    for (int i = o * k; i <= n; i++) {
      if (s[i] == '-')
        ans -= ksm(a, n - i) * ksm(b, i);
      else
        ans += ksm(a, n - i) * ksm(b, i);
      ans %= mol;
    }
  }
  printf("%I64d", (ans + mol) % mol);
  return 0;
}

```