#include <bits/stdc++.h>
using namespace std;
int s[200010];
char st[200010];
long long ksm(long long a, long long b) {
  long long ans = 1, tmp = (a % 1000000009 + 1000000009) % 1000000009;
  while (b) {
    if (b & 1) ans = ans * tmp % 1000000009;
    tmp = tmp * tmp % 1000000009;
    b >>= 1;
  }
  return ans;
}
int main() {
  int n, k;
  long long a, b;
  scanf("%d%lld%lld%d", &n, &a, &b, &k);
  long long inva = ksm(a, 1000000009 - 2);
  long long invb = ksm(b, 1000000009 - 2);
  scanf("%s", st);
  for (int i = 0; i < k; i++)
    if (st[i] == '+')
      s[i] = 1;
    else
      s[i] = -1;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long bi = ksm(inva, k), S = 0;
    (bi *= ksm(b, k)) %= 1000000009;
    long long a1 = ksm(a, n - i) * ksm(b, i) % 1000000009;
    long long nn = (n + 1) / k;
    if (bi == 1)
      S = a1 * nn % 1000000009;
    else
      S = a1 * ksm(1 - bi, 1000000009 - 2) % 1000000009 *
          (1 - ksm(bi, nn) + 1000000009) % 1000000009;
    if (s[i] == -1) S = -S + 1000000009;
    ans = ((ans + S) % 1000000009 + 1000000009) % 1000000009;
  }
  printf("%lld\n", ans);
  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;
int s[200010];
char st[200010];
long long ksm(long long a, long long b) {
  long long ans = 1, tmp = (a % 1000000009 + 1000000009) % 1000000009;
  while (b) {
    if (b & 1) ans = ans * tmp % 1000000009;
    tmp = tmp * tmp % 1000000009;
    b >>= 1;
  }
  return ans;
}
int main() {
  int n, k;
  long long a, b;
  scanf("%d%lld%lld%d", &n, &a, &b, &k);
  long long inva = ksm(a, 1000000009 - 2);
  long long invb = ksm(b, 1000000009 - 2);
  scanf("%s", st);
  for (int i = 0; i < k; i++)
    if (st[i] == '+')
      s[i] = 1;
    else
      s[i] = -1;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long bi = ksm(inva, k), S = 0;
    (bi *= ksm(b, k)) %= 1000000009;
    long long a1 = ksm(a, n - i) * ksm(b, i) % 1000000009;
    long long nn = (n + 1) / k;
    if (bi == 1)
      S = a1 * nn % 1000000009;
    else
      S = a1 * ksm(1 - bi, 1000000009 - 2) % 1000000009 *
          (1 - ksm(bi, nn) + 1000000009) % 1000000009;
    if (s[i] == -1) S = -S + 1000000009;
    ans = ((ans + S) % 1000000009 + 1000000009) % 1000000009;
  }
  printf("%lld\n", ans);
  return 0;
}

```