#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-6;
const int maxn = 1e6 + 10;
const long long mod = 1e9 + 9;
long long qpower(long long a, long long b, long long md) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % md;
    b >>= 1;
    a = (a * a) % md;
  }
  return ans;
}
string s;
int main(int argc, char const *argv[]) {
  int n;
  int a, b, k;
  std::cin >> n >> a >> b >> k;
  std::cin >> s;
  long long now = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      now += qpower(a, n - i, mod) * qpower(b, i, mod) % mod;
    } else
      now -= qpower(a, n - i, mod) * qpower(b, i, mod) % mod;
    now = (now + mod) % mod;
  }
  long long t =
      qpower(a, k * (mod - 2LL) % (mod - 1), mod) * qpower(b, k, mod) % mod;
  long long res = 0;
  if (t == 1) {
    res = now * ((n + 1) / k) % mod;
  } else {
    res = now * qpower(t - 1, mod - 2, mod) % mod *
          (qpower(t, (n + 1) / k, mod) - 1) % mod;
  }
  std::cout << res << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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 double eps = 1e-6;
const int maxn = 1e6 + 10;
const long long mod = 1e9 + 9;
long long qpower(long long a, long long b, long long md) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % md;
    b >>= 1;
    a = (a * a) % md;
  }
  return ans;
}
string s;
int main(int argc, char const *argv[]) {
  int n;
  int a, b, k;
  std::cin >> n >> a >> b >> k;
  std::cin >> s;
  long long now = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      now += qpower(a, n - i, mod) * qpower(b, i, mod) % mod;
    } else
      now -= qpower(a, n - i, mod) * qpower(b, i, mod) % mod;
    now = (now + mod) % mod;
  }
  long long t =
      qpower(a, k * (mod - 2LL) % (mod - 1), mod) * qpower(b, k, mod) % mod;
  long long res = 0;
  if (t == 1) {
    res = now * ((n + 1) / k) % mod;
  } else {
    res = now * qpower(t - 1, mod - 2, mod) % mod *
          (qpower(t, (n + 1) / k, mod) - 1) % mod;
  }
  std::cout << res << '\n';
  return 0;
}

```