#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 9;
long long power(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x) % M;
    x = (x * x) % M;
    n >>= 1;
  }
  return res;
}
int main(int argc, char const *argv[]) {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long n1 = (n + 1) / k;
  string s;
  cin >> s;
  long long num, den;
  if (a > b) {
    num = (power(a, n + 1) - power(b, n + 1) + M) % M;
    den = ((power(a, k) - power(b, k) + M) % M * power(a, n + 1 - k)) % M;
  } else {
    num = (power(b, n + 1) - power(a, n + 1) + M) % M;
    den = ((power(b, k) - power(a, k) + M) % M * power(a, n + 1 - k)) % M;
  }
  num = (num * power(den, M - 2)) % M;
  long long p = (b * power(a, M - 2)) % M;
  p = power(p, k);
  long long c1 = 0, c2 = 0;
  for (long long i = 0; i < k; i++) {
    long long y = (power(a, n - i) * power(b, i)) % M;
    if (p == 1) {
      if (s[i] == '+')
        c1 += (y * n1) % M;
      else
        c2 += (y * n1) % M;
    } else {
      long long x = (y * num) % M;
      if (s[i] == '+')
        c1 += x;
      else
        c2 += x;
    }
  }
  long long res = ((c1 - c2) % M + M) % M;
  cout << res << endl;
  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 M = 1e9 + 9;
long long power(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x) % M;
    x = (x * x) % M;
    n >>= 1;
  }
  return res;
}
int main(int argc, char const *argv[]) {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long n1 = (n + 1) / k;
  string s;
  cin >> s;
  long long num, den;
  if (a > b) {
    num = (power(a, n + 1) - power(b, n + 1) + M) % M;
    den = ((power(a, k) - power(b, k) + M) % M * power(a, n + 1 - k)) % M;
  } else {
    num = (power(b, n + 1) - power(a, n + 1) + M) % M;
    den = ((power(b, k) - power(a, k) + M) % M * power(a, n + 1 - k)) % M;
  }
  num = (num * power(den, M - 2)) % M;
  long long p = (b * power(a, M - 2)) % M;
  p = power(p, k);
  long long c1 = 0, c2 = 0;
  for (long long i = 0; i < k; i++) {
    long long y = (power(a, n - i) * power(b, i)) % M;
    if (p == 1) {
      if (s[i] == '+')
        c1 += (y * n1) % M;
      else
        c2 += (y * n1) % M;
    } else {
      long long x = (y * num) % M;
      if (s[i] == '+')
        c1 += x;
      else
        c2 += x;
    }
  }
  long long res = ((c1 - c2) % M + M) % M;
  cout << res << endl;
  return 0;
}

```