#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 9;
long long int fpow(long long int b, int p) {
  long long int r = 1ll;
  while (p) {
    if (p & 1) r *= b, r %= P;
    b = 1ll * b * b % P;
    p >>= 1;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  long long int ans = 0ll;
  for (int i = 0; (0 >= k ? i > k : i < k); i += 1 - 2 * (0 >= k)) {
    char c;
    cin >> c;
    long long int r = 1ll * b * fpow(a, P - 2) % P;
    long long int d = fpow(r, k) % P;
    long long int ord = (n - i) / k + 1;
    long long int num = (fpow(d, ord) - 1) % P;
    long long int cnt = 1ll * num * fpow(d - 1, P - 2) % P;
    long long int term = 1ll * fpow(a, n - i) * fpow(b, i) % P;
    cnt = 1ll * cnt * term % P;
    if (d == 1) cnt = 1ll * ord * term % P;
    if (c == '+')
      ans = (ans + cnt) % P;
    else
      ans = (ans - cnt + P) % P;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please formulate 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;
const int P = 1e9 + 9;
long long int fpow(long long int b, int p) {
  long long int r = 1ll;
  while (p) {
    if (p & 1) r *= b, r %= P;
    b = 1ll * b * b % P;
    p >>= 1;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  long long int ans = 0ll;
  for (int i = 0; (0 >= k ? i > k : i < k); i += 1 - 2 * (0 >= k)) {
    char c;
    cin >> c;
    long long int r = 1ll * b * fpow(a, P - 2) % P;
    long long int d = fpow(r, k) % P;
    long long int ord = (n - i) / k + 1;
    long long int num = (fpow(d, ord) - 1) % P;
    long long int cnt = 1ll * num * fpow(d - 1, P - 2) % P;
    long long int term = 1ll * fpow(a, n - i) * fpow(b, i) % P;
    cnt = 1ll * cnt * term % P;
    if (d == 1) cnt = 1ll * ord * term % P;
    if (c == '+')
      ans = (ans + cnt) % P;
    else
      ans = (ans - cnt + P) % P;
  }
  cout << ans << '\n';
  return 0;
}

```