#include <bits/stdc++.h>
using namespace std;
int n, k;
long long a, b, sum, ans, summ;
char s[100010];
long long powe(long long x, int r) {
  long long ret = 1;
  while (r) {
    if (r & 1) ret = (ret * x) % 1000000009;
    x = x * x % 1000000009;
    r >>= 1;
  }
  return ret;
}
int main() {
  scanf("%d%lld%lld%d", &n, &a, &b, &k);
  scanf("%s", s);
  long long p = b * powe(a, 1000000009 - 2) % 1000000009, d = powe(a, n);
  for (int i = 0; i < k; ++i) {
    if ((n + 1) % k == i) summ = sum;
    if (s[i] == '+')
      sum = (sum + d) % 1000000009;
    else
      sum = (sum - d + 1000000009) % 1000000009;
    d = (d * p) % 1000000009;
  }
  p = powe(p, k);
  int bt = (n + 1) / k;
  if (p == 1)
    ans = sum * bt;
  else
    ans = sum * ((powe(p, bt) - 1 + 1000000009) % 1000000009) % 1000000009 *
          powe((p - 1 + 1000000009) % 1000000009, 1000000009 - 2) % 1000000009;
  summ = summ * powe(p, bt) % 1000000009;
  ans = ((ans + summ) % 1000000009 + 1000000009) % 1000000009;
  printf("%lld", ans);
  return 0;
}

### Prompt

Please formulate 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, k;
long long a, b, sum, ans, summ;
char s[100010];
long long powe(long long x, int r) {
  long long ret = 1;
  while (r) {
    if (r & 1) ret = (ret * x) % 1000000009;
    x = x * x % 1000000009;
    r >>= 1;
  }
  return ret;
}
int main() {
  scanf("%d%lld%lld%d", &n, &a, &b, &k);
  scanf("%s", s);
  long long p = b * powe(a, 1000000009 - 2) % 1000000009, d = powe(a, n);
  for (int i = 0; i < k; ++i) {
    if ((n + 1) % k == i) summ = sum;
    if (s[i] == '+')
      sum = (sum + d) % 1000000009;
    else
      sum = (sum - d + 1000000009) % 1000000009;
    d = (d * p) % 1000000009;
  }
  p = powe(p, k);
  int bt = (n + 1) / k;
  if (p == 1)
    ans = sum * bt;
  else
    ans = sum * ((powe(p, bt) - 1 + 1000000009) % 1000000009) % 1000000009 *
          powe((p - 1 + 1000000009) % 1000000009, 1000000009 - 2) % 1000000009;
  summ = summ * powe(p, bt) % 1000000009;
  ans = ((ans + summ) % 1000000009 + 1000000009) % 1000000009;
  printf("%lld", ans);
  return 0;
}

```