#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
const int Mod = 1e9 + 9;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) (ans *= a) %= Mod;
    (a *= a) %= Mod;
    b >>= 1;
  }
  return ans;
}
long long n, a, b, k;
string s;
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long a1 = 0;
  for (int i = 0; i < s.size(); i++) {
    long long flag = 1;
    if (s[i] == '-') flag = -1;
    a1 = (a1 + flag * (qpow(a, n - i) * qpow(b, i)) % Mod + Mod) % Mod;
  }
  long long m = (n + 1) / k;
  long long ak = qpow(a, k);
  long long bk = qpow(b, k);
  long long x = qpow(ak, Mod - 2);
  long long q = (bk * x) % Mod;
  long long ans = 0;
  if (q != 1) {
    ans = (a1 * ((qpow(q, m) - 1)) % Mod + Mod) % Mod;
    long long t1 = qpow(q - 1, Mod - 2);
    ans = (ans * t1) % Mod;
  } else
    ans = (a1 * m) % Mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 int maxn = 1e5 + 50;
const int Mod = 1e9 + 9;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) (ans *= a) %= Mod;
    (a *= a) %= Mod;
    b >>= 1;
  }
  return ans;
}
long long n, a, b, k;
string s;
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long a1 = 0;
  for (int i = 0; i < s.size(); i++) {
    long long flag = 1;
    if (s[i] == '-') flag = -1;
    a1 = (a1 + flag * (qpow(a, n - i) * qpow(b, i)) % Mod + Mod) % Mod;
  }
  long long m = (n + 1) / k;
  long long ak = qpow(a, k);
  long long bk = qpow(b, k);
  long long x = qpow(ak, Mod - 2);
  long long q = (bk * x) % Mod;
  long long ans = 0;
  if (q != 1) {
    ans = (a1 * ((qpow(q, m) - 1)) % Mod + Mod) % Mod;
    long long t1 = qpow(q - 1, Mod - 2);
    ans = (ans * t1) % Mod;
  } else
    ans = (a1 * m) % Mod;
  cout << ans << endl;
  return 0;
}

```