#include <bits/stdc++.h>
const long long maxn = int(2e5) + 7, mod = int(1e9) + 9;
long long n, a, b, pow_ab[maxn], k, sum;
long long base_ab, base_a, base_b, flag[maxn];
char str[maxn];
long long power_mod(long long p, long long q) {
  long long ret = 1;
  while (q) {
    if (q & 1) ret = ret * p % mod;
    p = p * p % mod;
    q >>= 1;
  }
  return ret;
}
void init() {
  pow_ab[0] = power_mod(a, n);
  base_b = b;
  base_a = power_mod(a, mod - 2);
  for (int i = 1; i < k; i++)
    pow_ab[i] = pow_ab[i - 1] * base_a % mod * base_b % mod;
  base_ab = power_mod(power_mod(a, k), mod - 2) * power_mod(b, k) % mod;
}
int main() {
  std::cin >> n >> a >> b >> k >> str;
  if (n + 1 < k) k = n + 1;
  init();
  for (int i = 0; i < k; i++) flag[i] = (str[i] == '-' ? -1 : 1);
  long long t = (n + 1) / k;
  if (base_ab == 1)
    for (int i = 0; i < k; i++)
      sum = (sum + (pow_ab[i] * flag[i] + mod) % mod * t % mod) % mod;
  else {
    long long tmp = (1ll - power_mod(base_ab, t) + mod) % mod;
    tmp = tmp * power_mod((1ll - base_ab + mod) % mod, mod - 2) % mod;
    for (int i = 0; i < k; i++)
      sum = (sum + (pow_ab[i] * flag[i] + mod) % mod * tmp % mod) % mod;
  }
  std::cout << sum % mod << std::endl;
  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>
const long long maxn = int(2e5) + 7, mod = int(1e9) + 9;
long long n, a, b, pow_ab[maxn], k, sum;
long long base_ab, base_a, base_b, flag[maxn];
char str[maxn];
long long power_mod(long long p, long long q) {
  long long ret = 1;
  while (q) {
    if (q & 1) ret = ret * p % mod;
    p = p * p % mod;
    q >>= 1;
  }
  return ret;
}
void init() {
  pow_ab[0] = power_mod(a, n);
  base_b = b;
  base_a = power_mod(a, mod - 2);
  for (int i = 1; i < k; i++)
    pow_ab[i] = pow_ab[i - 1] * base_a % mod * base_b % mod;
  base_ab = power_mod(power_mod(a, k), mod - 2) * power_mod(b, k) % mod;
}
int main() {
  std::cin >> n >> a >> b >> k >> str;
  if (n + 1 < k) k = n + 1;
  init();
  for (int i = 0; i < k; i++) flag[i] = (str[i] == '-' ? -1 : 1);
  long long t = (n + 1) / k;
  if (base_ab == 1)
    for (int i = 0; i < k; i++)
      sum = (sum + (pow_ab[i] * flag[i] + mod) % mod * t % mod) % mod;
  else {
    long long tmp = (1ll - power_mod(base_ab, t) + mod) % mod;
    tmp = tmp * power_mod((1ll - base_ab + mod) % mod, mod - 2) % mod;
    for (int i = 0; i < k; i++)
      sum = (sum + (pow_ab[i] * flag[i] + mod) % mod * tmp % mod) % mod;
  }
  std::cout << sum % mod << std::endl;
  return 0;
}

```