#include <bits/stdc++.h>
using namespace std;
int n, a, b, k, ans;
char s[100005];
int fe(int x, int y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  int res = fe(x, y / 2);
  res = 1ll * res * res % 1000000009;
  if (y & 1) res = 1ll * res * x % 1000000009;
  return res;
}
int main() {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", s + 1);
  int invval = fe(1ll * b * fe(a, 1000000009 - 2) % 1000000009, k);
  int va = fe(a, n);
  int inva = fe(a, 1000000009 - 2);
  int vb = 1;
  int t1 = 0;
  for (int i = 1; i <= k; i++) {
    t1 = (1ll * t1 +
          ((s[i] == '-' ? -1ll : 1ll) * va * vb % 1000000009 + 1000000009) %
              1000000009) %
         1000000009;
    va = 1ll * va * inva % 1000000009;
    vb = 1ll * vb * b % 1000000009;
  }
  if (invval == 1) {
    printf("%d", 1ll * t1 * (n + 1) / k % 1000000009);
    return 0;
  }
  ans = 1ll * (1ll * fe(invval, (n + 1) / k) - 1 + 1000000009) % 1000000009 *
        fe((invval - 1 + 1000000009) % 1000000009, 1000000009 - 2) %
        1000000009 * t1 % 1000000009;
  printf("%d", ans);
}

### Prompt

Your task is to create 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;
int n, a, b, k, ans;
char s[100005];
int fe(int x, int y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  int res = fe(x, y / 2);
  res = 1ll * res * res % 1000000009;
  if (y & 1) res = 1ll * res * x % 1000000009;
  return res;
}
int main() {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", s + 1);
  int invval = fe(1ll * b * fe(a, 1000000009 - 2) % 1000000009, k);
  int va = fe(a, n);
  int inva = fe(a, 1000000009 - 2);
  int vb = 1;
  int t1 = 0;
  for (int i = 1; i <= k; i++) {
    t1 = (1ll * t1 +
          ((s[i] == '-' ? -1ll : 1ll) * va * vb % 1000000009 + 1000000009) %
              1000000009) %
         1000000009;
    va = 1ll * va * inva % 1000000009;
    vb = 1ll * vb * b % 1000000009;
  }
  if (invval == 1) {
    printf("%d", 1ll * t1 * (n + 1) / k % 1000000009);
    return 0;
  }
  ans = 1ll * (1ll * fe(invval, (n + 1) / k) - 1 + 1000000009) % 1000000009 *
        fe((invval - 1 + 1000000009) % 1000000009, 1000000009 - 2) %
        1000000009 * t1 % 1000000009;
  printf("%d", ans);
}

```