#include <bits/stdc++.h>
using namespace std;
long long powmod(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans = ans * x % 1000000009;
    }
    x = x * x % 1000000009;
    n >>= 1;
  }
  return ans;
}
long long invmod(long long x) { return powmod(x, 1000000009 - 2); }
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  long long n, a, b, k;
  string s;
  cin >> n >> a >> b >> k;
  cin >> s;
  long long ans = 0;
  long long mult = powmod(b, k) * invmod(powmod(a, k)) % 1000000009;
  long long part = 0;
  long long y = 1;
  for (int i = 0; i < k; i++) {
    long long x = powmod(a, n - i);
    long long cur = x * y % 1000000009;
    if (s[i] == '-') {
      cur *= -1;
      cur = (cur % 1000000009 + 1000000009) % 1000000009;
    }
    part += cur;
    part %= 1000000009;
    y = y * b % 1000000009;
  }
  long long cnt = (n + 1) / k;
  ans = mult == 1
            ? cnt
            : (powmod(mult, cnt) - 1 + 1000000009) % 1000000009 *
                  invmod((mult - 1 + 1000000009) % 1000000009) % 1000000009;
  ans = ans * part % 1000000009;
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate 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 powmod(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans = ans * x % 1000000009;
    }
    x = x * x % 1000000009;
    n >>= 1;
  }
  return ans;
}
long long invmod(long long x) { return powmod(x, 1000000009 - 2); }
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  long long n, a, b, k;
  string s;
  cin >> n >> a >> b >> k;
  cin >> s;
  long long ans = 0;
  long long mult = powmod(b, k) * invmod(powmod(a, k)) % 1000000009;
  long long part = 0;
  long long y = 1;
  for (int i = 0; i < k; i++) {
    long long x = powmod(a, n - i);
    long long cur = x * y % 1000000009;
    if (s[i] == '-') {
      cur *= -1;
      cur = (cur % 1000000009 + 1000000009) % 1000000009;
    }
    part += cur;
    part %= 1000000009;
    y = y * b % 1000000009;
  }
  long long cnt = (n + 1) / k;
  ans = mult == 1
            ? cnt
            : (powmod(mult, cnt) - 1 + 1000000009) % 1000000009 *
                  invmod((mult - 1 + 1000000009) % 1000000009) % 1000000009;
  ans = ans * part % 1000000009;
  cout << ans << "\n";
  return 0;
}

```