#include <bits/stdc++.h>
using namespace std;
const int N = (int)(1e3 + 5);
inline long long pwm(long long a, long long p) {
  long long s = 1;
  while (p > 0)
    if (p & 1) {
      p--;
      s = (s * a) % 1000000009;
    } else {
      p >>= 1;
      a = (a * a) % 1000000009;
    }
  return s;
}
inline long long inv(long long a) {
  return pwm(a % 1000000009, 1000000009 - 2);
}
inline long long mul(long long a, long long b) {
  return ((a % 1000000009 * b % 1000000009) % 1000000009 + 1000000009) %
         1000000009;
}
inline long long sgn(char c) { return c == '+' ? 1 : -1; }
inline long long dvd(long long a, long long b) { return mul(a, inv(b)); }
long long n, a, b, k;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cin >> n >> a >> b >> k;
  cin >> s;
  long long m = (n + 1) / k;
  long long b_a = dvd(b, a);
  long long sum = 0;
  for (int i = 0; i < k; i++)
    sum = (sum + mul(sgn(s[i]), pwm(b_a, i))) % 1000000009;
  if (a != b) {
    long long z = pwm(b_a, k);
    long long res =
        z == 1 ? m : dvd(1000000009 - 1 + pwm(z, m), 1000000009 - 1 + z);
    m = mul(res, pwm(a, n));
  }
  cout << mul(sum, m) << endl;
  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;
const int N = (int)(1e3 + 5);
inline long long pwm(long long a, long long p) {
  long long s = 1;
  while (p > 0)
    if (p & 1) {
      p--;
      s = (s * a) % 1000000009;
    } else {
      p >>= 1;
      a = (a * a) % 1000000009;
    }
  return s;
}
inline long long inv(long long a) {
  return pwm(a % 1000000009, 1000000009 - 2);
}
inline long long mul(long long a, long long b) {
  return ((a % 1000000009 * b % 1000000009) % 1000000009 + 1000000009) %
         1000000009;
}
inline long long sgn(char c) { return c == '+' ? 1 : -1; }
inline long long dvd(long long a, long long b) { return mul(a, inv(b)); }
long long n, a, b, k;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cin >> n >> a >> b >> k;
  cin >> s;
  long long m = (n + 1) / k;
  long long b_a = dvd(b, a);
  long long sum = 0;
  for (int i = 0; i < k; i++)
    sum = (sum + mul(sgn(s[i]), pwm(b_a, i))) % 1000000009;
  if (a != b) {
    long long z = pwm(b_a, k);
    long long res =
        z == 1 ? m : dvd(1000000009 - 1 + pwm(z, m), 1000000009 - 1 + z);
    m = mul(res, pwm(a, n));
  }
  cout << mul(sum, m) << endl;
  return 0;
}

```