#include <bits/stdc++.h>
using namespace std;
const long long mod = (long long)1e9 + 9;
long long quick_pow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
long long x, y, ans, n, a, b, k;
char s;
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b)
    x = 1, y = 0;
  else
    exgcd(b, a % b, y, x), y -= a / b * x;
}
int main() {
  cin >> n >> a >> b >> k;
  for (long long i = 0, d; i < k; i++) {
    cin >> s;
    s == '+' ? d = 1 : d = -1;
    long long a1 = (quick_pow(a, n - i) * quick_pow(b, i) % mod + mod) % mod;
    ans = (ans + d * a1 + mod) % mod;
  }
  exgcd(a, mod, x, y);
  x = (x % mod + mod) % mod;
  long long q = quick_pow(b * x % mod, k), jie = (n + 1) / k;
  if (q == 1)
    ans = (ans * jie % mod + mod) % mod;
  else {
    ans = (ans * (1 - quick_pow(q, jie)) % mod + mod) % mod;
    exgcd(((1 - q) % mod + mod) % mod, mod, x, y);
    x = (x % mod + mod) % mod;
    ans = (ans * x % mod + mod) % mod;
  }
  cout << ans % mod;
}

### Prompt

Your task is to create a CPP solution to 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>
using namespace std;
const long long mod = (long long)1e9 + 9;
long long quick_pow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
long long x, y, ans, n, a, b, k;
char s;
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b)
    x = 1, y = 0;
  else
    exgcd(b, a % b, y, x), y -= a / b * x;
}
int main() {
  cin >> n >> a >> b >> k;
  for (long long i = 0, d; i < k; i++) {
    cin >> s;
    s == '+' ? d = 1 : d = -1;
    long long a1 = (quick_pow(a, n - i) * quick_pow(b, i) % mod + mod) % mod;
    ans = (ans + d * a1 + mod) % mod;
  }
  exgcd(a, mod, x, y);
  x = (x % mod + mod) % mod;
  long long q = quick_pow(b * x % mod, k), jie = (n + 1) / k;
  if (q == 1)
    ans = (ans * jie % mod + mod) % mod;
  else {
    ans = (ans * (1 - quick_pow(q, jie)) % mod + mod) % mod;
    exgcd(((1 - q) % mod + mod) % mod, mod, x, y);
    x = (x % mod + mod) % mod;
    ans = (ans * x % mod + mod) % mod;
  }
  cout << ans % mod;
}

```