#include <bits/stdc++.h>
using namespace std;
long long n, x, m, q, k;
long long mulmod(long long a, long long b) {
  long long ret = 0;
  while (b) {
    if (b & 1) ret = (ret + a) % 1000000009;
    a = (a << 1) % 1000000009;
    b >>= 1;
  }
  return ret;
}
long long fastPow(long long x, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = mulmod(ret, x);
    x = mulmod(x, x);
    n >>= 1;
  }
  return ret;
}
long long a, b;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> a >> b >> k;
  long long ans = 0;
  string str;
  cin >> str;
  long long inva = fastPow(a, 1000000009 - 2);
  long long aux = 0, term, q;
  q = b * inva % 1000000009;
  term = fastPow(a, n);
  for (int i = int(0); i < int(k); i++) {
    char c = str[i % k];
    if (c == '+') {
      aux += term;
    } else {
      aux += 1000000009;
      aux -= term;
    }
    aux %= 1000000009;
    term = mulmod(term, q);
  }
  q = fastPow(q, k);
  int quot = (n + 1) / k;
  long long inv = fastPow(q - 1 + 1000000009, 1000000009 - 2);
  if (q <= 1)
    q = q * quot;
  else
    q = mulmod(((fastPow(q, quot) - 1) % 1000000009), inv);
  ans = mulmod(aux, q);
  cout << ans << '\n';
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 n, x, m, q, k;
long long mulmod(long long a, long long b) {
  long long ret = 0;
  while (b) {
    if (b & 1) ret = (ret + a) % 1000000009;
    a = (a << 1) % 1000000009;
    b >>= 1;
  }
  return ret;
}
long long fastPow(long long x, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = mulmod(ret, x);
    x = mulmod(x, x);
    n >>= 1;
  }
  return ret;
}
long long a, b;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> a >> b >> k;
  long long ans = 0;
  string str;
  cin >> str;
  long long inva = fastPow(a, 1000000009 - 2);
  long long aux = 0, term, q;
  q = b * inva % 1000000009;
  term = fastPow(a, n);
  for (int i = int(0); i < int(k); i++) {
    char c = str[i % k];
    if (c == '+') {
      aux += term;
    } else {
      aux += 1000000009;
      aux -= term;
    }
    aux %= 1000000009;
    term = mulmod(term, q);
  }
  q = fastPow(q, k);
  int quot = (n + 1) / k;
  long long inv = fastPow(q - 1 + 1000000009, 1000000009 - 2);
  if (q <= 1)
    q = q * quot;
  else
    q = mulmod(((fastPow(q, quot) - 1) % 1000000009), inv);
  ans = mulmod(aux, q);
  cout << ans << '\n';
  return 0;
}

```