#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 9;
long long poww(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2 == 1) res = res * x % mod;
    x = x * x % mod;
    n = n / 2;
  }
  return res;
}
int main() {
  long long n, a, b, k;
  char s[100010];
  cin >> n >> a >> b >> k;
  for (int i = 0; i <= k - 1; i++) {
    cin >> s[i];
  }
  long long ans, ans1, ans2, ans3, ans22;
  ans1 = (poww(b, n + 1) * poww(a, (n + 1) * (mod - 2)) - 1) % mod;
  ans2 = poww(b, k) * poww(a, k * (mod - 2));
  ans22 = (ans2 - 1) % mod;
  ans = (ans1 * poww(ans22, mod - 2)) % mod;
  long long res = 0;
  for (int i = 0; i <= k - 1; i++) {
    if (s[i] == '+') {
      res += poww(a, n - i) * poww(b, i) % mod;
    }
    if (s[i] == '-') {
      res = (((res - poww(a, n - i) * poww(b, i) + mod) % mod) + mod) % mod;
    }
  }
  if (ans2 == 1) {
    ans3 = (res * (n + 1) / k) % mod;
  } else {
    ans3 = (res * ans) % mod;
  }
  cout << ans3;
}

### Prompt

Please create a solution in CPP 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;
long long mod = 1e9 + 9;
long long poww(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2 == 1) res = res * x % mod;
    x = x * x % mod;
    n = n / 2;
  }
  return res;
}
int main() {
  long long n, a, b, k;
  char s[100010];
  cin >> n >> a >> b >> k;
  for (int i = 0; i <= k - 1; i++) {
    cin >> s[i];
  }
  long long ans, ans1, ans2, ans3, ans22;
  ans1 = (poww(b, n + 1) * poww(a, (n + 1) * (mod - 2)) - 1) % mod;
  ans2 = poww(b, k) * poww(a, k * (mod - 2));
  ans22 = (ans2 - 1) % mod;
  ans = (ans1 * poww(ans22, mod - 2)) % mod;
  long long res = 0;
  for (int i = 0; i <= k - 1; i++) {
    if (s[i] == '+') {
      res += poww(a, n - i) * poww(b, i) % mod;
    }
    if (s[i] == '-') {
      res = (((res - poww(a, n - i) * poww(b, i) + mod) % mod) + mod) % mod;
    }
  }
  if (ans2 == 1) {
    ans3 = (res * (n + 1) / k) % mod;
  } else {
    ans3 = (res * ans) % mod;
  }
  cout << ans3;
}

```