#include <bits/stdc++.h>
using namespace std;
char str[1000005];
long long solve(long long x, int y) {
  long long gop = x % 1000000009;
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * gop) % 1000000009;
    gop = (gop * gop) % 1000000009;
    y >>= 1;
  }
  return ret;
}
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", str);
  long long ak = solve(a, k);
  long long bk = solve(b, k);
  long long sum = 0;
  for (int i = 0; i < k; i++) {
    long long pls = (solve(a, k - 1 - i) * solve(b, i)) % 1000000009;
    if (str[i] == '+')
      sum = (sum + pls) % 1000000009;
    else
      sum = (sum + 1000000009 - pls) % 1000000009;
  }
  if (ak == bk) {
    long long ans = (solve(a, n + 1 - k) * ((n + 1) / k)) % 1000000009;
    ans = (ans * sum) % 1000000009;
    printf("%lld\n", ans);
  } else {
    long long ans =
        (1000000009 + solve(a, n + 1) - solve(b, n + 1)) % 1000000009;
    ans = (ans * solve(ak + 1000000009 - bk, 1000000009 - 2)) % 1000000009;
    ans = (ans * sum) % 1000000009;
    printf("%lld\n", ans);
  }
}

### Prompt

Your task is to create 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;
char str[1000005];
long long solve(long long x, int y) {
  long long gop = x % 1000000009;
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * gop) % 1000000009;
    gop = (gop * gop) % 1000000009;
    y >>= 1;
  }
  return ret;
}
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", str);
  long long ak = solve(a, k);
  long long bk = solve(b, k);
  long long sum = 0;
  for (int i = 0; i < k; i++) {
    long long pls = (solve(a, k - 1 - i) * solve(b, i)) % 1000000009;
    if (str[i] == '+')
      sum = (sum + pls) % 1000000009;
    else
      sum = (sum + 1000000009 - pls) % 1000000009;
  }
  if (ak == bk) {
    long long ans = (solve(a, n + 1 - k) * ((n + 1) / k)) % 1000000009;
    ans = (ans * sum) % 1000000009;
    printf("%lld\n", ans);
  } else {
    long long ans =
        (1000000009 + solve(a, n + 1) - solve(b, n + 1)) % 1000000009;
    ans = (ans * solve(ak + 1000000009 - bk, 1000000009 - 2)) % 1000000009;
    ans = (ans * sum) % 1000000009;
    printf("%lld\n", ans);
  }
}

```