#include <bits/stdc++.h>
using namespace std;
const double g = 10.0, eps = 1e-12;
const int N = 100000 + 10, maxn = 200000 + 10, inf = 0x3f3f3f3f,
          INF = 0x3f3f3f3f3f3f3f3f;
long long quick(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000009;
    a = a * a % 1000000009;
    b >>= 1;
  }
  return ans;
}
char s[N];
int main() {
  long long n, a, b, k;
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  n++;
  scanf("%s", s + 1);
  long long sum = 0;
  for (int i = 1; i <= k; i++) {
    sum +=
        (s[i] == '+' ? 1 : -1) * quick(a, n - i) * quick(b, i - 1) % 1000000009;
    sum = (sum % 1000000009 + 1000000009) % 1000000009;
  }
  long long bs = quick(b * quick(a, 1000000009 - 2) % 1000000009, k);
  if (bs == 1) {
    sum = (sum * (long long)(n / k)) % 1000000009;
    return 0 * printf("%lld\n", sum);
  }
  sum = sum * (1 - quick(bs, n / k)) % 1000000009 *
        quick(1 - bs, 1000000009 - 2) % 1000000009;
  sum = (sum % 1000000009 + 1000000009) % 1000000009;
  printf("%lld\n", sum);
  return 0;
}

### 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;
const double g = 10.0, eps = 1e-12;
const int N = 100000 + 10, maxn = 200000 + 10, inf = 0x3f3f3f3f,
          INF = 0x3f3f3f3f3f3f3f3f;
long long quick(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000009;
    a = a * a % 1000000009;
    b >>= 1;
  }
  return ans;
}
char s[N];
int main() {
  long long n, a, b, k;
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  n++;
  scanf("%s", s + 1);
  long long sum = 0;
  for (int i = 1; i <= k; i++) {
    sum +=
        (s[i] == '+' ? 1 : -1) * quick(a, n - i) * quick(b, i - 1) % 1000000009;
    sum = (sum % 1000000009 + 1000000009) % 1000000009;
  }
  long long bs = quick(b * quick(a, 1000000009 - 2) % 1000000009, k);
  if (bs == 1) {
    sum = (sum * (long long)(n / k)) % 1000000009;
    return 0 * printf("%lld\n", sum);
  }
  sum = sum * (1 - quick(bs, n / k)) % 1000000009 *
        quick(1 - bs, 1000000009 - 2) % 1000000009;
  sum = (sum % 1000000009 + 1000000009) % 1000000009;
  printf("%lld\n", sum);
  return 0;
}

```