#include <bits/stdc++.h>
using namespace std;
using namespace std;
long long multi(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b % 2 == 1) {
      ans *= a;
      ans %= 1000000009LL;
    }
    a *= a;
    a %= 1000000009LL;
    b /= 2;
  }
  return ans;
}
int main() {
  iostream::sync_with_stdio(false);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long begin_sum = 0;
  long long hoge = multi(a, n);
  hoge %= 1000000009LL;
  long long bobo = multi(a, 1000000009LL - 2);
  for (long long i = 0; i < k; ++i) {
    if (s[i] == '+') {
      begin_sum += hoge;
    } else {
      begin_sum -= hoge;
    }
    begin_sum += 1000000009LL;
    begin_sum %= 1000000009LL;
    hoge *= bobo;
    hoge %= 1000000009LL;
    hoge *= b;
    hoge %= 1000000009LL;
  }
  long long geko = multi(bobo, k);
  geko *= multi(b, k);
  geko %= 1000000009LL;
  if (geko == 1) {
    long long ans = ((n + 1LL) / k);
    ans *= begin_sum;
    ans %= 1000000009LL;
    ans += (long long)(abs(ans) / 1000000009LL + 1LL) * 1000000009LL;
    ans %= 1000000009LL;
    cout << ans << endl;
    return 0;
  } else {
    long long bunbo = geko - 1LL;
    bunbo += 1000000009LL;
    bunbo %= 1000000009LL;
    long long bunsi = multi(geko, (n + 1) / k) - 1LL;
    bunsi += 1000000009LL;
    bunsi %= 1000000009LL;
    bunsi *= begin_sum;
    bunsi %= 1000000009LL;
    long long ans = bunsi * multi(bunbo, 1000000009LL - 2);
    ans %= 1000000009LL;
    cout << ans << endl;
    return 0;
  }
  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;
using namespace std;
long long multi(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b % 2 == 1) {
      ans *= a;
      ans %= 1000000009LL;
    }
    a *= a;
    a %= 1000000009LL;
    b /= 2;
  }
  return ans;
}
int main() {
  iostream::sync_with_stdio(false);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long begin_sum = 0;
  long long hoge = multi(a, n);
  hoge %= 1000000009LL;
  long long bobo = multi(a, 1000000009LL - 2);
  for (long long i = 0; i < k; ++i) {
    if (s[i] == '+') {
      begin_sum += hoge;
    } else {
      begin_sum -= hoge;
    }
    begin_sum += 1000000009LL;
    begin_sum %= 1000000009LL;
    hoge *= bobo;
    hoge %= 1000000009LL;
    hoge *= b;
    hoge %= 1000000009LL;
  }
  long long geko = multi(bobo, k);
  geko *= multi(b, k);
  geko %= 1000000009LL;
  if (geko == 1) {
    long long ans = ((n + 1LL) / k);
    ans *= begin_sum;
    ans %= 1000000009LL;
    ans += (long long)(abs(ans) / 1000000009LL + 1LL) * 1000000009LL;
    ans %= 1000000009LL;
    cout << ans << endl;
    return 0;
  } else {
    long long bunbo = geko - 1LL;
    bunbo += 1000000009LL;
    bunbo %= 1000000009LL;
    long long bunsi = multi(geko, (n + 1) / k) - 1LL;
    bunsi += 1000000009LL;
    bunsi %= 1000000009LL;
    bunsi *= begin_sum;
    bunsi %= 1000000009LL;
    long long ans = bunsi * multi(bunbo, 1000000009LL - 2);
    ans %= 1000000009LL;
    cout << ans << endl;
    return 0;
  }
  return 0;
}

```