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

### 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 a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
long long inv(long long a) { return pow_mod(a, MOD - 2); }
int main() {
  long long n, a, b, k;
  while (cin >> n >> a >> b >> k) {
    string s;
    cin >> s;
    long long fuck = pow_mod(b, k) * inv(pow_mod(a, k)) % MOD;
    long long csl = 0;
    for (long long i = 0; i < k; i++) {
      long long sgn = 1;
      if (s[i] == '-') sgn = -1;
      csl += sgn * pow_mod(a, n - i) * pow_mod(b, i);
      csl += MOD;
      csl %= MOD;
    }
    if (fuck == 1) {
      long long ans = csl * (n + 1) / k;
      ans = (ans % MOD + MOD) % MOD;
      cout << ans << endl;
    } else {
      long long ans = csl * (1 - pow_mod(fuck, (n + 1) / k)) % MOD;
      ans %= MOD;
      ans *= inv(1 - fuck);
      ans = (ans % MOD + MOD) % MOD;
      cout << ans << endl;
    }
  }
}

```