#include <bits/stdc++.h>
const int mod = 1e9 + 9;
using namespace std;
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  long long d = a;
  while (b) {
    if (b & 1) ans = (ans * d) % m;
    d = (d * d) % m;
    b = b >> 1;
  }
  return ans % m;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long z = 0;
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+') {
      z = (z + (power(a, n - i, mod) * power(b, i, mod)) % mod) % mod;
    } else {
      z = (z - (power(a, n - i, mod) * power(b, i, mod)) % mod) % mod;
    }
  }
  long long inv_a = power(a, mod - 2, mod);
  long long q = (power(b, k, mod) * power(inv_a, k, mod)) % mod;
  long long dd = (n + 1) / k;
  long long ans = 0;
  if (q == 1)
    ans = (dd * z) % mod;
  else {
    long long inv_q = power(q - 1, mod - 2, mod);
    ans = (((z * (power(q, dd, mod) - 1)) % mod) * inv_q) % mod;
  }
  q = power(q, dd, mod);
  if (ans < 0) ans += mod;
  ans = ans % mod;
  cout << ans << endl;
  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>
const int mod = 1e9 + 9;
using namespace std;
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  long long d = a;
  while (b) {
    if (b & 1) ans = (ans * d) % m;
    d = (d * d) % m;
    b = b >> 1;
  }
  return ans % m;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long z = 0;
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+') {
      z = (z + (power(a, n - i, mod) * power(b, i, mod)) % mod) % mod;
    } else {
      z = (z - (power(a, n - i, mod) * power(b, i, mod)) % mod) % mod;
    }
  }
  long long inv_a = power(a, mod - 2, mod);
  long long q = (power(b, k, mod) * power(inv_a, k, mod)) % mod;
  long long dd = (n + 1) / k;
  long long ans = 0;
  if (q == 1)
    ans = (dd * z) % mod;
  else {
    long long inv_q = power(q - 1, mod - 2, mod);
    ans = (((z * (power(q, dd, mod) - 1)) % mod) * inv_q) % mod;
  }
  q = power(q, dd, mod);
  if (ans < 0) ans += mod;
  ans = ans % mod;
  cout << ans << endl;
  return 0;
}

```