#include <bits/stdc++.h>
using namespace std;
const int nmax = 1e+6 + 3;
long long MIN = ((long long)1 << 62) - 1;
long long MAX = -MIN;
double eps = 1e-15;
long long mod = 1e+9 + 9;
long long bin_pow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = (res * x) % mod;
      --y;
    } else {
      x = (x * x) % mod;
      y /= 2;
    }
  }
  return res;
}
long long n, a, b, k;
void solve() {
  cin >> n >> a >> b >> k;
  char ch;
  long long ans = 0;
  for (int i = 0; i < k; ++i) {
    int p = 0;
    cin >> ch;
    if (ch == '+')
      ++p;
    else
      --p;
    long long cn = ((p * bin_pow(a, n - i) * bin_pow(b, i) + mod) % mod);
    ans = (ans + ((p * bin_pow(a, n - i) * bin_pow(b, i) + mod) % mod)) % mod;
  }
  long long cn = (n + 1) / k;
  long long q = (bin_pow(b, k) * bin_pow(bin_pow(a, k), mod - 2)) % mod;
  if (q == 1)
    ans = (cn * ans) % mod;
  else
    ans = (((ans * (bin_pow(q, cn) - 1 + mod)) % mod) *
           bin_pow((q - 1 + mod) % mod, mod - 2)) %
          mod;
  cout << (ans + mod) % mod;
}
int main() {
  int t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
const int nmax = 1e+6 + 3;
long long MIN = ((long long)1 << 62) - 1;
long long MAX = -MIN;
double eps = 1e-15;
long long mod = 1e+9 + 9;
long long bin_pow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) {
      res = (res * x) % mod;
      --y;
    } else {
      x = (x * x) % mod;
      y /= 2;
    }
  }
  return res;
}
long long n, a, b, k;
void solve() {
  cin >> n >> a >> b >> k;
  char ch;
  long long ans = 0;
  for (int i = 0; i < k; ++i) {
    int p = 0;
    cin >> ch;
    if (ch == '+')
      ++p;
    else
      --p;
    long long cn = ((p * bin_pow(a, n - i) * bin_pow(b, i) + mod) % mod);
    ans = (ans + ((p * bin_pow(a, n - i) * bin_pow(b, i) + mod) % mod)) % mod;
  }
  long long cn = (n + 1) / k;
  long long q = (bin_pow(b, k) * bin_pow(bin_pow(a, k), mod - 2)) % mod;
  if (q == 1)
    ans = (cn * ans) % mod;
  else
    ans = (((ans * (bin_pow(q, cn) - 1 + mod)) % mod) *
           bin_pow((q - 1 + mod) % mod, mod - 2)) %
          mod;
  cout << (ans + mod) % mod;
}
int main() {
  int t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}

```