#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long int MODN = 1e9 + 9;
inline long long int modpow(long long int x, long long int exp) {
  long long int rtn = 1;
  while (exp) {
    if (exp & 1) rtn = rtn * x % MODN;
    exp >>= 1, x = x * x % MODN;
  }
  return rtn;
}
inline long long int inv(long long int x) { return modpow(x, MODN - 2); }
long long int n, a, b, k;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> a >> b >> k >> s;
  assert((n + 1) % k == 0);
  assert(s.size() == k);
  long long int tmp = modpow(a, n), sum = 0, r = b * inv(a) % MODN;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+')
      sum = (sum + tmp) % MODN;
    else
      sum = (sum - tmp) % MODN;
    tmp = (tmp * r) % MODN;
  }
  r = modpow(r, k), n = (n + 1) / k;
  if (r == 1) {
    cout << ((sum * n % MODN) + MODN) % MODN << endl;
  } else {
    long long int q = inv(r - 1), p = (modpow(r, n) - 1) * sum % MODN;
    cout << ((p * q % MODN) + MODN) % MODN << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
#pragma GCC optimize("O3")
using namespace std;
const long long int MODN = 1e9 + 9;
inline long long int modpow(long long int x, long long int exp) {
  long long int rtn = 1;
  while (exp) {
    if (exp & 1) rtn = rtn * x % MODN;
    exp >>= 1, x = x * x % MODN;
  }
  return rtn;
}
inline long long int inv(long long int x) { return modpow(x, MODN - 2); }
long long int n, a, b, k;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> a >> b >> k >> s;
  assert((n + 1) % k == 0);
  assert(s.size() == k);
  long long int tmp = modpow(a, n), sum = 0, r = b * inv(a) % MODN;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+')
      sum = (sum + tmp) % MODN;
    else
      sum = (sum - tmp) % MODN;
    tmp = (tmp * r) % MODN;
  }
  r = modpow(r, k), n = (n + 1) / k;
  if (r == 1) {
    cout << ((sum * n % MODN) + MODN) % MODN << endl;
  } else {
    long long int q = inv(r - 1), p = (modpow(r, n) - 1) * sum % MODN;
    cout << ((p * q % MODN) + MODN) % MODN << endl;
  }
  return 0;
}

```