#include <bits/stdc++.h>
using namespace std;
long long int exp_mod(long long int a, long long int n,
                      long long int m = 1000000009) {
  if (n == 0) {
    return 1;
  } else if (n % 2 == 0) {
    return (exp_mod((a * a) % 1000000009, n / 2, 1000000009)) % 1000000009;
  } else
    return (a * (exp_mod((a * a) % 1000000009, n / 2))) % 1000000009;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, a, b, k, i, j;
  cin >> n >> a >> b >> k;
  int A[k];
  char str[100000];
  cin >> str;
  char ch;
  long long int sum1 = 0;
  long long int sum2 = 0;
  long long int ainv = exp_mod(a, 1000000009 - 2);
  long long int p = (ainv * b) % 1000000009;
  long long int score;
  score = exp_mod(a, n);
  if (str[0] == '+')
    sum1 = score;
  else
    sum2 = score;
  for (i = 1; i < k; i++) {
    score = (score * p) % 1000000009;
    ch = str[i];
    if (ch == '+') {
      sum1 += score;
      sum1 %= 1000000009;
    } else {
      sum2 += score;
      sum2 %= 1000000009;
    }
  }
  long long int sum = sum1 - sum2;
  long long int q = exp_mod(p, k);
  long long int pro;
  if (q != 1) {
    pro = 1;
    pro = exp_mod(q, (n + 1) / k);
    pro -= 1;
    long long int quot = q - 1;
    long long int quot_inv = exp_mod(quot, 1000000009 - 2);
    pro *= quot_inv;
    pro %= 1000000009;
  } else {
    pro = (n + 1) / k;
  }
  sum *= pro;
  sum %= 1000000009;
  if (sum < 0) {
    sum += 1000000009;
  }
  cout << sum << '\n';
  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;
long long int exp_mod(long long int a, long long int n,
                      long long int m = 1000000009) {
  if (n == 0) {
    return 1;
  } else if (n % 2 == 0) {
    return (exp_mod((a * a) % 1000000009, n / 2, 1000000009)) % 1000000009;
  } else
    return (a * (exp_mod((a * a) % 1000000009, n / 2))) % 1000000009;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, a, b, k, i, j;
  cin >> n >> a >> b >> k;
  int A[k];
  char str[100000];
  cin >> str;
  char ch;
  long long int sum1 = 0;
  long long int sum2 = 0;
  long long int ainv = exp_mod(a, 1000000009 - 2);
  long long int p = (ainv * b) % 1000000009;
  long long int score;
  score = exp_mod(a, n);
  if (str[0] == '+')
    sum1 = score;
  else
    sum2 = score;
  for (i = 1; i < k; i++) {
    score = (score * p) % 1000000009;
    ch = str[i];
    if (ch == '+') {
      sum1 += score;
      sum1 %= 1000000009;
    } else {
      sum2 += score;
      sum2 %= 1000000009;
    }
  }
  long long int sum = sum1 - sum2;
  long long int q = exp_mod(p, k);
  long long int pro;
  if (q != 1) {
    pro = 1;
    pro = exp_mod(q, (n + 1) / k);
    pro -= 1;
    long long int quot = q - 1;
    long long int quot_inv = exp_mod(quot, 1000000009 - 2);
    pro *= quot_inv;
    pro %= 1000000009;
  } else {
    pro = (n + 1) / k;
  }
  sum *= pro;
  sum %= 1000000009;
  if (sum < 0) {
    sum += 1000000009;
  }
  cout << sum << '\n';
  return 0;
}

```