#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 9;
char s[100010];
long long inv, n, a, b, k, q, len, a1, fz;
long long qmi(long long x, long long y) {
  long long tmp = x, ans = 1;
  while (y) {
    if (y & 1) ans = (ans * tmp) % mod;
    tmp = (tmp * tmp) % mod;
    y >>= 1;
  }
  return ans;
}
int main() {
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  long long ans = 0;
  inv = qmi(a, mod - 2);
  q = qmi(inv, k) * qmi(b, k) % mod;
  len = (n + 1) / k;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+')
      a1 = qmi(a, n - i) * qmi(b, i) % mod;
    else
      a1 = (-1ll * (qmi(a, n - i) * qmi(b, i)) % mod + mod) % mod;
    if (qmi(a, k) == qmi(b, k)) {
      ans = (ans + a1 * len % mod) % mod;
    } else {
      fz = (qmi(q - 1, mod - 2) + mod) % mod;
      ans = (ans + (a1 * (qmi(q, len) - 1) % mod + mod) % mod * fz % mod) % mod;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 mod = 1e9 + 9;
char s[100010];
long long inv, n, a, b, k, q, len, a1, fz;
long long qmi(long long x, long long y) {
  long long tmp = x, ans = 1;
  while (y) {
    if (y & 1) ans = (ans * tmp) % mod;
    tmp = (tmp * tmp) % mod;
    y >>= 1;
  }
  return ans;
}
int main() {
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  long long ans = 0;
  inv = qmi(a, mod - 2);
  q = qmi(inv, k) * qmi(b, k) % mod;
  len = (n + 1) / k;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+')
      a1 = qmi(a, n - i) * qmi(b, i) % mod;
    else
      a1 = (-1ll * (qmi(a, n - i) * qmi(b, i)) % mod + mod) % mod;
    if (qmi(a, k) == qmi(b, k)) {
      ans = (ans + a1 * len % mod) % mod;
    } else {
      fz = (qmi(q - 1, mod - 2) + mod) % mod;
      ans = (ans + (a1 * (qmi(q, len) - 1) % mod + mod) % mod * fz % mod) % mod;
    }
  }
  cout << ans;
  return 0;
}

```