#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 29;
const int MOD = 1e9 + 9;
const int N = 1e5 + 10;
char s[N];
long long power_mod(long long a, int n) {
  long long ret = 1;
  for (; n; n >>= 1) {
    if (n & 1) ret = ret * a % MOD;
    a = a * a % MOD;
  }
  return ret;
}
long long inv_mod(long long a) { return power_mod(a, MOD - 2); }
int main() {
  int n, A, B, m;
  scanf("%d%d%d%d", &n, &A, &B, &m);
  int q = inv_mod(A) * B % MOD;
  int sum = 0, prod = 1;
  scanf("%s", s);
  for (int i = 0; i < m; ++i) {
    if (s[i] == '+') {
      sum = (sum + prod) % MOD;
    } else {
      sum = (sum + MOD - prod) % MOD;
    }
    prod = (long long)prod * q % MOD;
  }
  q = power_mod(q, m);
  int ret;
  if (q == 1) {
    ret = (n + 1) / m;
  } else {
    ret = inv_mod(q - 1) * (power_mod(q, (n + 1) / m) + MOD - 1) % MOD;
  }
  ret = ret * power_mod(A, n) % MOD * sum % MOD;
  printf("%d\n", ret);
  return 0;
}

### 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int INF = 1 << 29;
const int MOD = 1e9 + 9;
const int N = 1e5 + 10;
char s[N];
long long power_mod(long long a, int n) {
  long long ret = 1;
  for (; n; n >>= 1) {
    if (n & 1) ret = ret * a % MOD;
    a = a * a % MOD;
  }
  return ret;
}
long long inv_mod(long long a) { return power_mod(a, MOD - 2); }
int main() {
  int n, A, B, m;
  scanf("%d%d%d%d", &n, &A, &B, &m);
  int q = inv_mod(A) * B % MOD;
  int sum = 0, prod = 1;
  scanf("%s", s);
  for (int i = 0; i < m; ++i) {
    if (s[i] == '+') {
      sum = (sum + prod) % MOD;
    } else {
      sum = (sum + MOD - prod) % MOD;
    }
    prod = (long long)prod * q % MOD;
  }
  q = power_mod(q, m);
  int ret;
  if (q == 1) {
    ret = (n + 1) / m;
  } else {
    ret = inv_mod(q - 1) * (power_mod(q, (n + 1) / m) + MOD - 1) % MOD;
  }
  ret = ret * power_mod(A, n) % MOD * sum % MOD;
  printf("%d\n", ret);
  return 0;
}

```