#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k, S, ans, md = 1e9 + 9;
string s;
long long pw(long long base, long long pr) {
  long long ret = 1;
  while (pr) {
    if ((pr & 1)) ret = (ret * base) % md;
    base = (base * base) % md;
    pr /= 2;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> a >> b >> k;
  cin >> s;
  long long d = pw((b * pw(a, md - 2)) % md, k);
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+')
      S = 1;
    else
      S = -1;
    long long fi = pw(a, n - i) * pw(b, i) % md;
    if (d == 1) {
      ans += S * (fi * (n / k + 1) % md);
      ans += md;
      ans %= md;
      continue;
    }
    long long tmp1 = (pw(d, n / k + 1) - 1 + md) % md;
    long long tmp2 = (pw((d - 1 + md) % md, md - 2) + md) % md;
    long long mul = tmp1 * tmp2 % md;
    ans += S * fi * mul % md;
    ans = ans + md;
    ans %= md;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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, S, ans, md = 1e9 + 9;
string s;
long long pw(long long base, long long pr) {
  long long ret = 1;
  while (pr) {
    if ((pr & 1)) ret = (ret * base) % md;
    base = (base * base) % md;
    pr /= 2;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> a >> b >> k;
  cin >> s;
  long long d = pw((b * pw(a, md - 2)) % md, k);
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+')
      S = 1;
    else
      S = -1;
    long long fi = pw(a, n - i) * pw(b, i) % md;
    if (d == 1) {
      ans += S * (fi * (n / k + 1) % md);
      ans += md;
      ans %= md;
      continue;
    }
    long long tmp1 = (pw(d, n / k + 1) - 1 + md) % md;
    long long tmp2 = (pw((d - 1 + md) % md, md - 2) + md) % md;
    long long mul = tmp1 * tmp2 % md;
    ans += S * fi * mul % md;
    ans = ans + md;
    ans %= md;
  }
  cout << ans << endl;
  return 0;
}

```