#include <bits/stdc++.h>
using namespace std;
long long M = (long long)(1e9 + 9);
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y % 2 == 1) {
      res = (res * x) % M;
    }
    y /= 2;
    x = (x * x) % M;
  }
  return res;
}
long long mod_inv(long long a) { return power(a, M - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a, b, n, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long ans = 0;
  long long x = power(a, n);
  long long y = (b * mod_inv(a)) % M;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      ans = (ans + x) % M;
    } else {
      ans = (ans - x + M) % M;
    }
    x = (x * y) % M;
  }
  x = power(y, k);
  n++;
  n /= k;
  if (x == 1) {
    ans = (ans * (n)) % M;
  } else {
    ans = (ans * ((power(x, n) - 1 + M) % M)) % M;
    ans = (ans * mod_inv((x - 1 + M) % M)) % M;
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 M = (long long)(1e9 + 9);
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y % 2 == 1) {
      res = (res * x) % M;
    }
    y /= 2;
    x = (x * x) % M;
  }
  return res;
}
long long mod_inv(long long a) { return power(a, M - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long a, b, n, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long ans = 0;
  long long x = power(a, n);
  long long y = (b * mod_inv(a)) % M;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      ans = (ans + x) % M;
    } else {
      ans = (ans - x + M) % M;
    }
    x = (x * y) % M;
  }
  x = power(y, k);
  n++;
  n /= k;
  if (x == 1) {
    ans = (ans * (n)) % M;
  } else {
    ans = (ans * ((power(x, n) - 1 + M) % M)) % M;
    ans = (ans * mod_inv((x - 1 + M) % M)) % M;
  }
  cout << ans << endl;
}

```