#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 9;
const long long MAX_N = 1e5 + 10;
long long qc(long long a, long long b) {
  long long ans = a;
  long long r = 1;
  while (b) {
    if (b & 1) r = (r * ans) % M;
    b >>= 1;
    ans = (ans * ans) % M;
  }
  return r;
}
int main() {
  long long n, a, b, t;
  char nums[MAX_N];
  scanf("%lld%lld%lld%lld", &n, &a, &b, &t);
  getchar();
  gets(nums);
  long long i = 0;
  long long SUM = 0;
  long long QQ = 0;
  for (int i = 0; i < t; i++) {
    QQ = (qc(a, n - i) * qc(b, i)) % M;
    if (nums[i] == '+') {
      SUM = ((SUM % M) + (QQ % M)) % M;
    } else {
      SUM = ((SUM % M) - (QQ % M)) % M;
    }
  }
  long long Number = n + 1;
  long long cnt = Number / t;
  long long QB = (b * qc(a, M - 2)) % M;
  QB = qc(QB, t);
  long long up, down;
  up = (SUM * ((qc(QB, cnt) - 1) % M)) % M;
  down = (QB - 1) % M;
  if (QB == 1) {
    SUM = (SUM * cnt) % M;
  } else {
    SUM = (up * qc(down, M - 2)) % M;
  }
  printf("%lld\n", (SUM + M) % M);
  return 0;
}

### Prompt

Your challenge is to write 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 long long M = 1e9 + 9;
const long long MAX_N = 1e5 + 10;
long long qc(long long a, long long b) {
  long long ans = a;
  long long r = 1;
  while (b) {
    if (b & 1) r = (r * ans) % M;
    b >>= 1;
    ans = (ans * ans) % M;
  }
  return r;
}
int main() {
  long long n, a, b, t;
  char nums[MAX_N];
  scanf("%lld%lld%lld%lld", &n, &a, &b, &t);
  getchar();
  gets(nums);
  long long i = 0;
  long long SUM = 0;
  long long QQ = 0;
  for (int i = 0; i < t; i++) {
    QQ = (qc(a, n - i) * qc(b, i)) % M;
    if (nums[i] == '+') {
      SUM = ((SUM % M) + (QQ % M)) % M;
    } else {
      SUM = ((SUM % M) - (QQ % M)) % M;
    }
  }
  long long Number = n + 1;
  long long cnt = Number / t;
  long long QB = (b * qc(a, M - 2)) % M;
  QB = qc(QB, t);
  long long up, down;
  up = (SUM * ((qc(QB, cnt) - 1) % M)) % M;
  down = (QB - 1) % M;
  if (QB == 1) {
    SUM = (SUM * cnt) % M;
  } else {
    SUM = (up * qc(down, M - 2)) % M;
  }
  printf("%lld\n", (SUM + M) % M);
  return 0;
}

```