#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 9;
long long qpow(long long x, long long n, long long m) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % m;
    n = n >> 1;
    x = x * x % m;
  }
  return res;
}
long long inv(long long x, long long m) { return qpow(x, m - 2, m); }
int main(void) {
  int n, a, b, k, i;
  cin >> n >> a >> b >> k;
  cin.get();
  long long gotcha = 0, q, resultado;
  for (i = 0; i < k; ++i) {
    char c;
    scanf("%c", &c);
    if (c == '+') {
      gotcha =
          (gotcha + qpow(a, n - i, mod) * qpow(b, i, mod) % mod + mod) % mod;
    } else {
      gotcha =
          (gotcha - qpow(a, n - i, mod) * qpow(b, i, mod) % mod + mod) % mod;
    }
  }
  q = qpow(b, k, mod) * inv(qpow(a, k, mod), mod) % mod;
  if (q == 1) {
    resultado = gotcha * (n + 1) / k % mod;
  } else {
    resultado =
        gotcha * (qpow(q, (n + 1) / k, mod) - 1) % mod * inv(q - 1, mod) % mod;
  }
  cout << resultado;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 int mod = 1e9 + 9;
long long qpow(long long x, long long n, long long m) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % m;
    n = n >> 1;
    x = x * x % m;
  }
  return res;
}
long long inv(long long x, long long m) { return qpow(x, m - 2, m); }
int main(void) {
  int n, a, b, k, i;
  cin >> n >> a >> b >> k;
  cin.get();
  long long gotcha = 0, q, resultado;
  for (i = 0; i < k; ++i) {
    char c;
    scanf("%c", &c);
    if (c == '+') {
      gotcha =
          (gotcha + qpow(a, n - i, mod) * qpow(b, i, mod) % mod + mod) % mod;
    } else {
      gotcha =
          (gotcha - qpow(a, n - i, mod) * qpow(b, i, mod) % mod + mod) % mod;
    }
  }
  q = qpow(b, k, mod) * inv(qpow(a, k, mod), mod) % mod;
  if (q == 1) {
    resultado = gotcha * (n + 1) / k % mod;
  } else {
    resultado =
        gotcha * (qpow(q, (n + 1) / k, mod) - 1) % mod * inv(q - 1, mod) % mod;
  }
  cout << resultado;
  return 0;
}

```