#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k, rs;
string s;
long long pw(long long a, long long p) {
  long long rs = 1;
  while (p) {
    if (p & 1) rs = (rs * a) % 1000000009;
    p >>= 1;
    a = (a * a) % 1000000009;
  }
  return rs % 1000000009;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> a >> b >> k;
  cin >> s;
  long long ratio = pw(pw(a, k), 1000000009 - 2) * pw(b, k) % 1000000009;
  long long aux = 0;
  for (int i = 0; i < k; i++) {
    int sign = (s[i] == '+' ? 1 : -1);
    long long fst = pw(a, n - i) * pw(b, i) % 1000000009;
    aux += (sign * fst + 1000000009) % 1000000009;
    aux %= 1000000009;
  }
  if (ratio == 1) return cout << (aux * ((n + 1) / k)) % 1000000009, 0;
  long long ans = aux;
  long long ft = pw(ratio, (long long)((n + 1) / k)) - 1;
  ft %= 1000000009;
  long long scd = pw(ratio - 1, 1000000009 - 2);
  scd %= 1000000009;
  ans = ans * ft % 1000000009;
  ans = ans * scd % 1000000009;
  rs += (ans + 1000000009) % 1000000009;
  rs %= 1000000009;
  while (rs < 0) rs += 1000000009;
  cout << (rs % 1000000009);
  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;
long long n, a, b, k, rs;
string s;
long long pw(long long a, long long p) {
  long long rs = 1;
  while (p) {
    if (p & 1) rs = (rs * a) % 1000000009;
    p >>= 1;
    a = (a * a) % 1000000009;
  }
  return rs % 1000000009;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> a >> b >> k;
  cin >> s;
  long long ratio = pw(pw(a, k), 1000000009 - 2) * pw(b, k) % 1000000009;
  long long aux = 0;
  for (int i = 0; i < k; i++) {
    int sign = (s[i] == '+' ? 1 : -1);
    long long fst = pw(a, n - i) * pw(b, i) % 1000000009;
    aux += (sign * fst + 1000000009) % 1000000009;
    aux %= 1000000009;
  }
  if (ratio == 1) return cout << (aux * ((n + 1) / k)) % 1000000009, 0;
  long long ans = aux;
  long long ft = pw(ratio, (long long)((n + 1) / k)) - 1;
  ft %= 1000000009;
  long long scd = pw(ratio - 1, 1000000009 - 2);
  scd %= 1000000009;
  ans = ans * ft % 1000000009;
  ans = ans * scd % 1000000009;
  rs += (ans + 1000000009) % 1000000009;
  rs %= 1000000009;
  while (rs < 0) rs += 1000000009;
  cout << (rs % 1000000009);
  return 0;
}

```