#include <bits/stdc++.h>
using std::pair;
using std::vector;
const int64_t mod = 1000 * 1000 * 1000 + 9;
int64_t Mult(int64_t x, int64_t y) { return (x * y) % mod; }
int64_t Add(int64_t x, int64_t y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
  return x;
}
int64_t Sub(int64_t x, int64_t y) {
  x -= y;
  if (x < 0) {
    x += mod;
  }
  return x;
}
int64_t BinPow(int64_t a, int64_t n) {
  if (n == 0) {
    return 1;
  }
  if (n & 1) {
    return Mult(a, BinPow(a, n - 1));
  }
  return BinPow(Mult(a, a), n / 2) % mod;
}
int64_t Rev(int64_t x) { return BinPow(x, mod - 2); }
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int64_t n, a, b, k;
  std::cin >> n >> a >> b >> k;
  vector<char> data(k);
  for (auto& elem : data) {
    std::cin >> elem;
  }
  int64_t cur = BinPow(a, n), c = Mult(b, Rev(a)), ans = 0;
  for (size_t i = 0; i < k; ++i) {
    if (data[i] == '+') {
      ans = Add(ans, cur);
    } else {
      ans = Sub(ans, cur);
    }
    cur = Mult(cur, c);
  }
  c = BinPow(c, k);
  int64_t d = (n + 1) / k;
  if (c == 1) {
    ans = Mult(ans, d);
  } else {
    ans = Mult(ans, Sub(BinPow(c, d), 1));
    ans = Mult(ans, Rev(Sub(c, 1)));
  }
  std::cout << ans;
  return 0;
}

### Prompt

Please formulate 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 std::pair;
using std::vector;
const int64_t mod = 1000 * 1000 * 1000 + 9;
int64_t Mult(int64_t x, int64_t y) { return (x * y) % mod; }
int64_t Add(int64_t x, int64_t y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
  return x;
}
int64_t Sub(int64_t x, int64_t y) {
  x -= y;
  if (x < 0) {
    x += mod;
  }
  return x;
}
int64_t BinPow(int64_t a, int64_t n) {
  if (n == 0) {
    return 1;
  }
  if (n & 1) {
    return Mult(a, BinPow(a, n - 1));
  }
  return BinPow(Mult(a, a), n / 2) % mod;
}
int64_t Rev(int64_t x) { return BinPow(x, mod - 2); }
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int64_t n, a, b, k;
  std::cin >> n >> a >> b >> k;
  vector<char> data(k);
  for (auto& elem : data) {
    std::cin >> elem;
  }
  int64_t cur = BinPow(a, n), c = Mult(b, Rev(a)), ans = 0;
  for (size_t i = 0; i < k; ++i) {
    if (data[i] == '+') {
      ans = Add(ans, cur);
    } else {
      ans = Sub(ans, cur);
    }
    cur = Mult(cur, c);
  }
  c = BinPow(c, k);
  int64_t d = (n + 1) / k;
  if (c == 1) {
    ans = Mult(ans, d);
  } else {
    ans = Mult(ans, Sub(BinPow(c, d), 1));
    ans = Mult(ans, Rev(Sub(c, 1)));
  }
  std::cout << ans;
  return 0;
}

```