#include <bits/stdc++.h>
using namespace std;
char s[100001];
long long ModExp(long long a, long long n) {
  a = (1000000009 + a % 1000000009) % 1000000009;
  long long ans = 1;
  a %= 1000000009;
  while (n) {
    if (n % 2) ans = (ans * a) % 1000000009;
    a = (a * a) % 1000000009;
    n /= 2;
  }
  return ans;
}
long long ModInv(long long a) {
  a = (1000000009 + a % 1000000009) % 1000000009;
  return ModExp(a, 1000000009 - 2);
}
int main() {
  long long n, a, b, k, m;
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  m = (n + 1) / k;
  long long S = 0, inva = ModInv(a), r = (b * inva) % 1000000009;
  for (long long i = 0, t = 1; i < k; ++i, t = (t * r) % 1000000009) {
    if (s[i] == '+')
      S += t;
    else
      S -= t;
    S = (1000000009 + S) % 1000000009;
  }
  S = (S * ModExp(a, n)) % 1000000009;
  r = ModExp(r, k);
  if (r == 1)
    S = (m * S) % 1000000009;
  else {
    S = (S * (ModExp(r, m) + 1000000009 - 1)) % 1000000009;
    S = (S * (ModInv(1000000009 + r - 1))) % 1000000009;
  }
  cout << S << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
char s[100001];
long long ModExp(long long a, long long n) {
  a = (1000000009 + a % 1000000009) % 1000000009;
  long long ans = 1;
  a %= 1000000009;
  while (n) {
    if (n % 2) ans = (ans * a) % 1000000009;
    a = (a * a) % 1000000009;
    n /= 2;
  }
  return ans;
}
long long ModInv(long long a) {
  a = (1000000009 + a % 1000000009) % 1000000009;
  return ModExp(a, 1000000009 - 2);
}
int main() {
  long long n, a, b, k, m;
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  m = (n + 1) / k;
  long long S = 0, inva = ModInv(a), r = (b * inva) % 1000000009;
  for (long long i = 0, t = 1; i < k; ++i, t = (t * r) % 1000000009) {
    if (s[i] == '+')
      S += t;
    else
      S -= t;
    S = (1000000009 + S) % 1000000009;
  }
  S = (S * ModExp(a, n)) % 1000000009;
  r = ModExp(r, k);
  if (r == 1)
    S = (m * S) % 1000000009;
  else {
    S = (S * (ModExp(r, m) + 1000000009 - 1)) % 1000000009;
    S = (S * (ModInv(1000000009 + r - 1))) % 1000000009;
  }
  cout << S << '\n';
  return 0;
}

```