#include <bits/stdc++.h>
using namespace std;
long long pow(long long x, long long p) {
  if (p == 0) return 1;
  if (p % 2) return x * pow(x, p - 1) % 1000000009;
  long long y = pow(x, p / 2);
  return y * y % 1000000009;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long coef[k];
  for (int i = 0; i < k; i++) {
    char c;
    cin >> c;
    coef[i] = (c == '+' ? 1 : -1);
  }
  long long ainv = pow(a, 1000000009 - 2), t = pow(a, n), ans = 0,
            diff = ainv * b % 1000000009;
  for (int i = 0; i < k; i++) {
    ans += coef[i] * t % 1000000009;
    t *= diff;
    t %= 1000000009;
    ans %= 1000000009;
    if (ans < 0) ans += 1000000009;
  }
  diff = pow(diff, k);
  if (diff == 1) {
    cout << ans * (n + 1) / k % 1000000009 << endl;
    return 0;
  }
  ans *= (pow(diff, (n + 1) / k) - 1) % 1000000009;
  ans %= 1000000009;
  ans *= pow(diff - 1, 1000000009 - 2);
  ans %= 1000000009;
  if (ans < 0) ans += 1000000009;
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 pow(long long x, long long p) {
  if (p == 0) return 1;
  if (p % 2) return x * pow(x, p - 1) % 1000000009;
  long long y = pow(x, p / 2);
  return y * y % 1000000009;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long coef[k];
  for (int i = 0; i < k; i++) {
    char c;
    cin >> c;
    coef[i] = (c == '+' ? 1 : -1);
  }
  long long ainv = pow(a, 1000000009 - 2), t = pow(a, n), ans = 0,
            diff = ainv * b % 1000000009;
  for (int i = 0; i < k; i++) {
    ans += coef[i] * t % 1000000009;
    t *= diff;
    t %= 1000000009;
    ans %= 1000000009;
    if (ans < 0) ans += 1000000009;
  }
  diff = pow(diff, k);
  if (diff == 1) {
    cout << ans * (n + 1) / k % 1000000009 << endl;
    return 0;
  }
  ans *= (pow(diff, (n + 1) / k) - 1) % 1000000009;
  ans %= 1000000009;
  ans *= pow(diff - 1, 1000000009 - 2);
  ans %= 1000000009;
  if (ans < 0) ans += 1000000009;
  cout << ans << endl;
  return 0;
}

```