#include <bits/stdc++.h>
using namespace std;
int n, a, b, k;
const int mod = 1e9 + 9;
char str[110000];
int power(int a, int n) {
  if (n == 0) return 1;
  int ret = power(a, n / 2);
  ret = 1LL * ret * ret % mod;
  if (n % 2) ret = 1LL * ret * a % mod;
  return ret;
}
int main() {
  int ans = 0, t, i, j, s, vl, p, q, pr;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  pr = power(1LL * b * power(a, 1LL * (mod - 2)) % mod, k);
  scanf("%s", str);
  for (i = 0; i < k; i++) {
    if (str[i] == '+')
      t = 1;
    else
      t = mod - 1;
    t = 1LL * t * power(a, n - i) % mod * power(b, i) % mod;
    if (pr == 1)
      ans = (ans + 1LL * t * ((n - i) / k + 1)) % mod;
    else {
      vl =
          1LL * (1 - power(pr, (n - i) / k + 1)) * power(1 - pr, mod - 2) % mod;
      ans = (ans + 1LL * t * vl) % mod;
    }
    if (ans < 0) ans += mod;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int mod = 1e9 + 9;
char str[110000];
int power(int a, int n) {
  if (n == 0) return 1;
  int ret = power(a, n / 2);
  ret = 1LL * ret * ret % mod;
  if (n % 2) ret = 1LL * ret * a % mod;
  return ret;
}
int main() {
  int ans = 0, t, i, j, s, vl, p, q, pr;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  pr = power(1LL * b * power(a, 1LL * (mod - 2)) % mod, k);
  scanf("%s", str);
  for (i = 0; i < k; i++) {
    if (str[i] == '+')
      t = 1;
    else
      t = mod - 1;
    t = 1LL * t * power(a, n - i) % mod * power(b, i) % mod;
    if (pr == 1)
      ans = (ans + 1LL * t * ((n - i) / k + 1)) % mod;
    else {
      vl =
          1LL * (1 - power(pr, (n - i) / k + 1)) * power(1 - pr, mod - 2) % mod;
      ans = (ans + 1LL * t * vl) % mod;
    }
    if (ans < 0) ans += mod;
  }
  printf("%d\n", ans);
  return 0;
}

```