#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000009;
long long power(long long x, long long y) {
  long long result = 1;
  if (x == 0) return 0;
  while (y) {
    if (y & 1) result = result * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return result;
}
string s;
long long temp = 0, vara, varb, inva, invb, r;
int main(int argc, char const *argv[]) {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> s;
  vara = power(a, n);
  varb = 1;
  inva = power(a, mod - 2);
  r = inva * b % mod;
  long long num = (n + 1) / k;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      temp = (temp + vara * varb % mod) % mod;
    else
      temp = (temp + (mod - 1) * vara % mod * varb % mod) % mod;
    vara = vara * inva % mod;
    varb = varb * b % mod;
  }
  r = power(r, k);
  if (r == 1) {
    cout << temp * num % mod << "\n";
    return 0;
  }
  long long hold1 = (power(r, num) - 1 + mod) % mod;
  long long hold2 = power(r - 1, mod - 2) % mod;
  long long sum = (hold1 * hold2) % mod;
  temp = (temp * sum) % mod;
  printf("%lld\n", temp);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 long long mod = 1000000009;
long long power(long long x, long long y) {
  long long result = 1;
  if (x == 0) return 0;
  while (y) {
    if (y & 1) result = result * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return result;
}
string s;
long long temp = 0, vara, varb, inva, invb, r;
int main(int argc, char const *argv[]) {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> s;
  vara = power(a, n);
  varb = 1;
  inva = power(a, mod - 2);
  r = inva * b % mod;
  long long num = (n + 1) / k;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      temp = (temp + vara * varb % mod) % mod;
    else
      temp = (temp + (mod - 1) * vara % mod * varb % mod) % mod;
    vara = vara * inva % mod;
    varb = varb * b % mod;
  }
  r = power(r, k);
  if (r == 1) {
    cout << temp * num % mod << "\n";
    return 0;
  }
  long long hold1 = (power(r, num) - 1 + mod) % mod;
  long long hold2 = power(r - 1, mod - 2) % mod;
  long long sum = (hold1 * hold2) % mod;
  temp = (temp * sum) % mod;
  printf("%lld\n", temp);
  return 0;
}

```