#include <bits/stdc++.h>
using namespace std;
int n, a, b, k;
long long pang(long long base, int p) {
  int now = 1;
  long long ret = 1;
  long long kali = base;
  while (now <= p) {
    if ((now & p) > 0) {
      ret = (ret * kali) % 1000000009;
    }
    kali = (kali * kali) % 1000000009;
    now = now * 2;
  }
  return ret;
}
int main() {
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  n++;
  long long x = pang(b, k);
  long long y = pang(a, k);
  y = pang(y, 1000000009 - 2);
  x = (x * y) % 1000000009;
  long long ini;
  if (x == 1) {
    ini = n / k;
  } else {
    long long r = x;
    long long bawah = pang(r - 1, 1000000009 - 2);
    long long atas = pang(r, n / k) - 1;
    ini = (atas * bawah) % 1000000009;
  }
  long long now = pang(a, n - 1);
  long long nex = b;
  long long kali = pang(a, 1000000009 - 2);
  nex = (nex * kali) % 1000000009;
  string baca;
  cin >> baca;
  long long jawab = 0;
  for (int i = 0; i < k; i++) {
    long long tamb = (now * ini) % 1000000009;
    if (baca[i] == '+') {
      jawab = (jawab + tamb) % 1000000009;
    } else {
      jawab = (jawab - tamb) % 1000000009;
      jawab = (jawab + 1000000009) % 1000000009;
    }
    now = (now * nex) % 1000000009;
  }
  cout << jawab << "\n";
}

### 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;
int n, a, b, k;
long long pang(long long base, int p) {
  int now = 1;
  long long ret = 1;
  long long kali = base;
  while (now <= p) {
    if ((now & p) > 0) {
      ret = (ret * kali) % 1000000009;
    }
    kali = (kali * kali) % 1000000009;
    now = now * 2;
  }
  return ret;
}
int main() {
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  n++;
  long long x = pang(b, k);
  long long y = pang(a, k);
  y = pang(y, 1000000009 - 2);
  x = (x * y) % 1000000009;
  long long ini;
  if (x == 1) {
    ini = n / k;
  } else {
    long long r = x;
    long long bawah = pang(r - 1, 1000000009 - 2);
    long long atas = pang(r, n / k) - 1;
    ini = (atas * bawah) % 1000000009;
  }
  long long now = pang(a, n - 1);
  long long nex = b;
  long long kali = pang(a, 1000000009 - 2);
  nex = (nex * kali) % 1000000009;
  string baca;
  cin >> baca;
  long long jawab = 0;
  for (int i = 0; i < k; i++) {
    long long tamb = (now * ini) % 1000000009;
    if (baca[i] == '+') {
      jawab = (jawab + tamb) % 1000000009;
    } else {
      jawab = (jawab - tamb) % 1000000009;
      jawab = (jawab + 1000000009) % 1000000009;
    }
    now = (now * nex) % 1000000009;
  }
  cout << jawab << "\n";
}

```