#include <bits/stdc++.h>
using namespace std;
long long int mod = 1000000009;
long long int power(long long int x, long long int y) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2) % mod;
  p = (p * p) % mod;
  return (y % 2 == 0) ? p : (x * p) % mod;
}
int main() {
  long long int n, a, b, k, arr[100001], i, sum = 0, temp, res;
  char t[100002];
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", t);
  for (i = 0; i < k; i++) {
    if (t[i] == '+')
      arr[i] = 1;
    else
      arr[i] = -1;
  }
  for (i = 0; i < k; i++) {
    sum += arr[i] * ((power(a, n - i) * power(b, i)) % mod);
    sum = sum % mod;
  }
  temp = (power(a, k * (mod - 2)) * power(b, k)) % mod;
  if (temp == 1)
    res = (sum * (n + 1) / k) % mod;
  else {
    res = (sum *
           (power(temp - 1, mod - 2) * (power(temp, (n + 1) / k) - 1) % mod)) %
          mod;
  }
  if (res < 0) res += mod;
  printf("%lld", res);
  return 0;
}

### 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;
long long int mod = 1000000009;
long long int power(long long int x, long long int y) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2) % mod;
  p = (p * p) % mod;
  return (y % 2 == 0) ? p : (x * p) % mod;
}
int main() {
  long long int n, a, b, k, arr[100001], i, sum = 0, temp, res;
  char t[100002];
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", t);
  for (i = 0; i < k; i++) {
    if (t[i] == '+')
      arr[i] = 1;
    else
      arr[i] = -1;
  }
  for (i = 0; i < k; i++) {
    sum += arr[i] * ((power(a, n - i) * power(b, i)) % mod);
    sum = sum % mod;
  }
  temp = (power(a, k * (mod - 2)) * power(b, k)) % mod;
  if (temp == 1)
    res = (sum * (n + 1) / k) % mod;
  else {
    res = (sum *
           (power(temp - 1, mod - 2) * (power(temp, (n + 1) / k) - 1) % mod)) %
          mod;
  }
  if (res < 0) res += mod;
  printf("%lld", res);
  return 0;
}

```