#include <bits/stdc++.h>
using namespace std;
long long mpow(long long a, long long b) {
  long long ans = 1LL;
  while (b) {
    if (b & 1) {
      ans = (ans % 1000000009 * a % 1000000009) % 1000000009;
    }
    a = (a % 1000000009 * a % 1000000009) % 1000000009;
    b /= 2;
  }
  return ans;
}
int main(int argc, char const *argv[]) {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long cur = 0;
  for (int i = 0; i < k; i++) {
    long long apow = mpow(a, n - i);
    long long bpow = mpow(b, i);
    long long tmp = (apow * bpow) % 1000000009;
    if (s[i] == '+')
      cur += tmp;
    else
      cur += 1000000009 - tmp;
    cur %= 1000000009;
  }
  long long fac = (n + 1) / k;
  long long n1 = mpow(b, n + 1);
  long long n2 = mpow(mpow(a, n + 1), 1000000009 - 2);
  long long tmp1 = (n1 * n2) % 1000000009;
  tmp1 = (tmp1 + 1000000009 - 1) % 1000000009;
  n1 = mpow(b, k);
  n2 = mpow(mpow(a, k), 1000000009 - 2);
  long long tmp2 = (n1 * n2) % 1000000009;
  tmp2 = (tmp2 + 1000000009 - 1) % 1000000009;
  bool flag = false;
  if (tmp2 != 0) flag = true;
  tmp2 = mpow(tmp2, 1000000009 - 2);
  long long ans;
  if (flag)
    ans = (tmp1 * tmp2) % 1000000009;
  else
    ans = fac;
  ans = (ans * cur) % 1000000009;
  cout << ans << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
long long mpow(long long a, long long b) {
  long long ans = 1LL;
  while (b) {
    if (b & 1) {
      ans = (ans % 1000000009 * a % 1000000009) % 1000000009;
    }
    a = (a % 1000000009 * a % 1000000009) % 1000000009;
    b /= 2;
  }
  return ans;
}
int main(int argc, char const *argv[]) {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long cur = 0;
  for (int i = 0; i < k; i++) {
    long long apow = mpow(a, n - i);
    long long bpow = mpow(b, i);
    long long tmp = (apow * bpow) % 1000000009;
    if (s[i] == '+')
      cur += tmp;
    else
      cur += 1000000009 - tmp;
    cur %= 1000000009;
  }
  long long fac = (n + 1) / k;
  long long n1 = mpow(b, n + 1);
  long long n2 = mpow(mpow(a, n + 1), 1000000009 - 2);
  long long tmp1 = (n1 * n2) % 1000000009;
  tmp1 = (tmp1 + 1000000009 - 1) % 1000000009;
  n1 = mpow(b, k);
  n2 = mpow(mpow(a, k), 1000000009 - 2);
  long long tmp2 = (n1 * n2) % 1000000009;
  tmp2 = (tmp2 + 1000000009 - 1) % 1000000009;
  bool flag = false;
  if (tmp2 != 0) flag = true;
  tmp2 = mpow(tmp2, 1000000009 - 2);
  long long ans;
  if (flag)
    ans = (tmp1 * tmp2) % 1000000009;
  else
    ans = fac;
  ans = (ans * cur) % 1000000009;
  cout << ans << "\n";
  return 0;
}

```