#include <bits/stdc++.h>
using namespace std;
char sign;
long long a, b, inf = 1e9 + 9, sum, n, k, tmp, cnt;
long long fp(long long base, long long power) {
  long long result = 1;
  while (power) {
    if (power & 1) {
      result = result * base % inf;
    }
    power >>= 1;
    base = (base * base) % inf;
  }
  return result;
}
int main() {
  sum = 0;
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%c", &sign);
  for (int i = 0; i < k; i++) {
    scanf("%c", &sign);
    if (sign == '+') {
      sum = (sum + (fp(a, n - i) * fp(b, i) % inf)) % inf;
    } else {
      sum = (sum + inf - (fp(a, n - i) * fp(b, i) % inf)) % inf;
    }
  }
  tmp = fp(b, k) * fp(fp(a, k), inf - 2) % inf;
  if (tmp == 1) {
    sum = sum * ((n + 1) / k) % inf;
    printf("%lld", sum);
  } else {
    sum =
        ((sum * (fp(tmp, (n + 1) / k) - 1)) % inf) * fp(tmp - 1, inf - 2) % inf;
    printf("%lld", sum);
  }
}

### Prompt

Create a solution in CPP for 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;
char sign;
long long a, b, inf = 1e9 + 9, sum, n, k, tmp, cnt;
long long fp(long long base, long long power) {
  long long result = 1;
  while (power) {
    if (power & 1) {
      result = result * base % inf;
    }
    power >>= 1;
    base = (base * base) % inf;
  }
  return result;
}
int main() {
  sum = 0;
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%c", &sign);
  for (int i = 0; i < k; i++) {
    scanf("%c", &sign);
    if (sign == '+') {
      sum = (sum + (fp(a, n - i) * fp(b, i) % inf)) % inf;
    } else {
      sum = (sum + inf - (fp(a, n - i) * fp(b, i) % inf)) % inf;
    }
  }
  tmp = fp(b, k) * fp(fp(a, k), inf - 2) % inf;
  if (tmp == 1) {
    sum = sum * ((n + 1) / k) % inf;
    printf("%lld", sum);
  } else {
    sum =
        ((sum * (fp(tmp, (n + 1) / k) - 1)) % inf) * fp(tmp - 1, inf - 2) % inf;
    printf("%lld", sum);
  }
}

```