#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 9;
long long pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b % 2 == 1) {
      ret *= a;
      ret %= MOD;
    }
    a *= a;
    a %= MOD;
    b = b >> 1;
  }
  return ret;
}
int main() {
  long long n, k;
  long long a, b;
  cin >> n >> a >> b >> k;
  n += 1;
  string s;
  cin >> s;
  long long ans = 0;
  for (long long i = 0; i < k; i++) {
    long long x = pow(a, n - i - 1);
    x *= pow(b, i);
    x %= MOD;
    if (s[i] == '+')
      ans += x;
    else
      ans -= x;
    ans %= MOD;
  }
  long long p = pow(b, k) * pow(pow(a, k), MOD - 2) % MOD;
  long long t = n / k;
  if (p == 1)
    ans *= t;
  else {
    long long y = pow(p, t) - 1;
    y += MOD;
    y %= MOD;
    y *= pow(p - 1, MOD - 2);
    y %= MOD;
    ans *= y;
  }
  ans %= MOD;
  ans += MOD;
  ans %= MOD;
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 MOD = 1e9 + 9;
long long pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b % 2 == 1) {
      ret *= a;
      ret %= MOD;
    }
    a *= a;
    a %= MOD;
    b = b >> 1;
  }
  return ret;
}
int main() {
  long long n, k;
  long long a, b;
  cin >> n >> a >> b >> k;
  n += 1;
  string s;
  cin >> s;
  long long ans = 0;
  for (long long i = 0; i < k; i++) {
    long long x = pow(a, n - i - 1);
    x *= pow(b, i);
    x %= MOD;
    if (s[i] == '+')
      ans += x;
    else
      ans -= x;
    ans %= MOD;
  }
  long long p = pow(b, k) * pow(pow(a, k), MOD - 2) % MOD;
  long long t = n / k;
  if (p == 1)
    ans *= t;
  else {
    long long y = pow(p, t) - 1;
    y += MOD;
    y %= MOD;
    y *= pow(p - 1, MOD - 2);
    y %= MOD;
    ans *= y;
  }
  ans %= MOD;
  ans += MOD;
  ans %= MOD;
  cout << ans << endl;
  return 0;
}

```