#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 9;
long long add(long long a, long long b) {
  return (((a + b) % mod) + mod) % mod;
}
long long mod_pow(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long sub = mod_pow(a, b / 2);
  if (b % 2 == 0) {
    return (sub * sub) % mod;
  } else {
    return (((sub * sub) % mod) * a) % mod;
  }
}
long long mul(long long a, long long b) { return (a * b) % mod; }
long long inv(long long a) { return mod_pow(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, k, sum = 0;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long fac = 1, inva = inv(a), initial = mod_pow(a, n);
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+') {
      sum = add(sum, mul(initial, fac));
    } else {
      sum = add(sum, -mul(initial, fac));
    }
    fac = mul(mul(fac, inva), b);
  }
  int r = (n + 1) / k;
  long long something;
  if (fac == 1) {
    something = r;
  } else {
    something = mul(add(mod_pow(fac, r), -1), inv((fac - 1 + mod) % mod));
  }
  cout << mul(sum, something) << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 add(long long a, long long b) {
  return (((a + b) % mod) + mod) % mod;
}
long long mod_pow(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long sub = mod_pow(a, b / 2);
  if (b % 2 == 0) {
    return (sub * sub) % mod;
  } else {
    return (((sub * sub) % mod) * a) % mod;
  }
}
long long mul(long long a, long long b) { return (a * b) % mod; }
long long inv(long long a) { return mod_pow(a, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, k, sum = 0;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long fac = 1, inva = inv(a), initial = mod_pow(a, n);
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+') {
      sum = add(sum, mul(initial, fac));
    } else {
      sum = add(sum, -mul(initial, fac));
    }
    fac = mul(mul(fac, inva), b);
  }
  int r = (n + 1) / k;
  long long something;
  if (fac == 1) {
    something = r;
  } else {
    something = mul(add(mod_pow(fac, r), -1), inv((fac - 1 + mod) % mod));
  }
  cout << mul(sum, something) << endl;
  return 0;
}

```