#include <bits/stdc++.h>
using namespace std;
long long int powermod(long long int b, long long int p, long long int m) {
  long long int r = 1;
  b %= m;
  while (p > 0) {
    if (p & 1) r = (r * b) % m;
    p >>= 1;
    b = (b * b) % m;
  }
  return r;
}
long long int m(long long int s) {
  return (((s % (1000000000 + 9)) + (1000000000 + 9)) % (1000000000 + 9));
}
vector<bool> str;
long long int N, A, B, K;
map<pair<long long int, long long int>, long long int> mp;
long long int compute(long long int s, long long int len) {
  if (mp[{s, len}] != 0) return mp[{s, len}];
  long long int total = 0;
  if (len < 300000) {
    for (int i = s, j = 0; j < len; ++i, ++j) {
      long long int w = (str[i % K]) ? 1 : -1;
      total = m(total + w * m(powermod(A, len - j - 1, (1000000000 + 9)) *
                              powermod(B, j, (1000000000 + 9))));
    }
  } else {
    long long int newlen = (((len / 2) / K) * K);
    long long int a =
        m(powermod(A, len - newlen, (1000000000 + 9)) * compute(s, newlen));
    long long int b = m(m(powermod(A, len - 2 * newlen, (1000000000 + 9)) *
                          powermod(B, newlen, (1000000000 + 9))) *
                        compute(s, newlen));
    long long int c = m(powermod(B, 2 * newlen, (1000000000 + 9)) *
                        compute(s + 2 * newlen, len - 2 * newlen));
    total = m(total + a + b + c);
  }
  return mp[{s, len}] = total;
}
int main() {
  cin >> N >> A >> B >> K;
  str = vector<bool>(K);
  char c;
  for (long long int i = (0); i < (long long int)(K); ++i) {
    scanf(" %c", &c);
    str[i] = c == '+';
  }
  cout << compute(0, N + 1) << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 powermod(long long int b, long long int p, long long int m) {
  long long int r = 1;
  b %= m;
  while (p > 0) {
    if (p & 1) r = (r * b) % m;
    p >>= 1;
    b = (b * b) % m;
  }
  return r;
}
long long int m(long long int s) {
  return (((s % (1000000000 + 9)) + (1000000000 + 9)) % (1000000000 + 9));
}
vector<bool> str;
long long int N, A, B, K;
map<pair<long long int, long long int>, long long int> mp;
long long int compute(long long int s, long long int len) {
  if (mp[{s, len}] != 0) return mp[{s, len}];
  long long int total = 0;
  if (len < 300000) {
    for (int i = s, j = 0; j < len; ++i, ++j) {
      long long int w = (str[i % K]) ? 1 : -1;
      total = m(total + w * m(powermod(A, len - j - 1, (1000000000 + 9)) *
                              powermod(B, j, (1000000000 + 9))));
    }
  } else {
    long long int newlen = (((len / 2) / K) * K);
    long long int a =
        m(powermod(A, len - newlen, (1000000000 + 9)) * compute(s, newlen));
    long long int b = m(m(powermod(A, len - 2 * newlen, (1000000000 + 9)) *
                          powermod(B, newlen, (1000000000 + 9))) *
                        compute(s, newlen));
    long long int c = m(powermod(B, 2 * newlen, (1000000000 + 9)) *
                        compute(s + 2 * newlen, len - 2 * newlen));
    total = m(total + a + b + c);
  }
  return mp[{s, len}] = total;
}
int main() {
  cin >> N >> A >> B >> K;
  str = vector<bool>(K);
  char c;
  for (long long int i = (0); i < (long long int)(K); ++i) {
    scanf(" %c", &c);
    str[i] = c == '+';
  }
  cout << compute(0, N + 1) << endl;
  return 0;
}

```