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

### Prompt

Develop a solution in CPP 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;
const long long mod = 1e9 + 9;
long long pow_mod(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    n >>= 1;
    x = x * x % mod;
  }
  return res;
}
long long inv(long long x) { return pow_mod(x, mod - 2); }
int main() {
  long long n, a, b, k;
  string s;
  while (cin >> n >> a >> b >> k) {
    cin >> s;
    long long ans = 0;
    for (long long i = 0; i < k; i++) {
      if (s[i] == '+')
        ans = (ans + pow_mod(a, n - i) * pow_mod(b, i) + mod) % mod;
      else
        ans = (ans - pow_mod(a, n - i) * pow_mod(b, i) + mod) % mod;
    }
    long long m = (n + 1) / k;
    long long q = pow_mod(b, k) * inv(pow_mod(a, k)) % mod;
    if (q == 1)
      ans = ans * m % mod;
    else {
      ans = ans * (1 - pow_mod(q, m)) % mod;
      ans = ans * inv(1 - q) % mod;
    }
    cout << (ans + mod) % mod << endl;
  }
  return 0;
}

```