#include <bits/stdc++.h>
using namespace std;
const int MODULO = 1000000009;
int power(int a, int b, int modulo) {
  if (b == 0) return 1 % modulo;
  if (b & 1) return 1LL * power(a, b - 1, modulo) * a % modulo;
  return power(1LL * a * a % modulo, b >> 1, modulo);
}
int inverse(int a, int modulo) { return power(a, modulo - 2, modulo); }
int powerSum(int base, int k) {
  if (k == 0) return 1 % MODULO;
  if (k & 1) {
    int extra = power(base, (k + 1) >> 1, MODULO);
    int half = powerSum(base, k >> 1);
    return 1LL * (extra + 1) * half % MODULO;
  } else
    return (powerSum(base, k - 1) + power(base, k, MODULO)) % MODULO;
}
int N, a, b, k;
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> a >> b >> k;
  cin >> s;
  int p = 0;
  for (int i = 0; i < k; i++) {
    int current = 1LL * power(a, N - i, MODULO) * power(b, i, MODULO) % MODULO;
    if (s[i] == '+')
      p = (p + current) % MODULO;
    else
      p = (p - current + MODULO) % MODULO;
  }
  int base =
      1LL * inverse(power(a, k, MODULO), MODULO) * power(b, k, MODULO) % MODULO;
  int s = powerSum(base, (N + 1) / k - 1);
  cout << 1LL * p * s % MODULO << endl;
  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;
const int MODULO = 1000000009;
int power(int a, int b, int modulo) {
  if (b == 0) return 1 % modulo;
  if (b & 1) return 1LL * power(a, b - 1, modulo) * a % modulo;
  return power(1LL * a * a % modulo, b >> 1, modulo);
}
int inverse(int a, int modulo) { return power(a, modulo - 2, modulo); }
int powerSum(int base, int k) {
  if (k == 0) return 1 % MODULO;
  if (k & 1) {
    int extra = power(base, (k + 1) >> 1, MODULO);
    int half = powerSum(base, k >> 1);
    return 1LL * (extra + 1) * half % MODULO;
  } else
    return (powerSum(base, k - 1) + power(base, k, MODULO)) % MODULO;
}
int N, a, b, k;
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> a >> b >> k;
  cin >> s;
  int p = 0;
  for (int i = 0; i < k; i++) {
    int current = 1LL * power(a, N - i, MODULO) * power(b, i, MODULO) % MODULO;
    if (s[i] == '+')
      p = (p + current) % MODULO;
    else
      p = (p - current + MODULO) % MODULO;
  }
  int base =
      1LL * inverse(power(a, k, MODULO), MODULO) * power(b, k, MODULO) % MODULO;
  int s = powerSum(base, (N + 1) / k - 1);
  cout << 1LL * p * s % MODULO << endl;
  return 0;
}

```