#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream& operator<<(ostream& os, vector<T> v) {
  for (long long i = (0); i < ((long long)(v).size()); i++)
    os << (i ? " " : "") << v[i];
  return os;
}
const bool debug = 1;
long long fpow(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long x = fpow(a, b / 2);
    return (x * x) % 1000000009;
  } else {
    return (a * fpow(a, b - 1)) % 1000000009;
  }
}
long long sig(char c) {
  if (c == '+')
    return 1;
  else
    return 1000000009 - 1;
}
signed main() {
  long long a, b, n, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long ans = 0;
  long long inva = fpow(a, 1000000009 - 2);
  long long r = fpow((b * inva) % 1000000009, k);
  long long x;
  if (r != 1)
    x = ((1 - fpow(r, (n + 1) / k)) * fpow(1 - r, 1000000009 - 2)) % 1000000009;
  else
    x = ((n + 1) / k % 1000000009);
  for (long long i = 0; i < k; i++) {
    ans =
        (ans + (sig(s[i]) *
                (x * (fpow(b, i) * fpow(a, n - i) % 1000000009) % 1000000009) %
                1000000009) %
                   1000000009) %
        1000000009;
  }
  cout << (ans + 1000000009) % 1000000009 << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
template <class T>
ostream& operator<<(ostream& os, vector<T> v) {
  for (long long i = (0); i < ((long long)(v).size()); i++)
    os << (i ? " " : "") << v[i];
  return os;
}
const bool debug = 1;
long long fpow(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long x = fpow(a, b / 2);
    return (x * x) % 1000000009;
  } else {
    return (a * fpow(a, b - 1)) % 1000000009;
  }
}
long long sig(char c) {
  if (c == '+')
    return 1;
  else
    return 1000000009 - 1;
}
signed main() {
  long long a, b, n, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long ans = 0;
  long long inva = fpow(a, 1000000009 - 2);
  long long r = fpow((b * inva) % 1000000009, k);
  long long x;
  if (r != 1)
    x = ((1 - fpow(r, (n + 1) / k)) * fpow(1 - r, 1000000009 - 2)) % 1000000009;
  else
    x = ((n + 1) / k % 1000000009);
  for (long long i = 0; i < k; i++) {
    ans =
        (ans + (sig(s[i]) *
                (x * (fpow(b, i) * fpow(a, n - i) % 1000000009) % 1000000009) %
                1000000009) %
                   1000000009) %
        1000000009;
  }
  cout << (ans + 1000000009) % 1000000009 << endl;
  return 0;
}

```