#include <bits/stdc++.h>
using namespace std;
long long mod = (int)1e9 + 9;
long long modPow(long long x, long long k) {
  long long res = 1;
  for (x %= mod; k; k >>= 1) {
    if (k & 1) res = (res * x) % mod;
    x = (x * x) % mod;
  }
  return res;
}
long long modInv(long long x) { return modPow(x, mod - 2); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  k = min(k, n + 1);
  vector<long long> s(k);
  for (int i = 0; i < k; ++i) {
    char c;
    cin >> c;
    s[i] = c == '+' ? 1 : mod - 1;
  }
  while ((k << 1) < min(n, (long long)1e5)) {
    vector<long long> temp(s.begin(), s.end());
    s.insert(s.end(), temp.begin(), temp.end());
    k <<= 1;
  }
  long long dp = 0;
  for (int i = 0; i < k; ++i) {
    long long p = s[i];
    p = (p * modPow(a, n - i)) % mod;
    p = (p * modPow(b, i)) % mod;
    dp = (dp + p) % mod;
  }
  long long res = 0;
  for (long long i = 0; i <= n; i += k) {
    if (i + k <= n + 1) {
      res = (res + dp * modPow(b * modInv(a), i)) % mod;
    } else {
      for (; i <= n; ++i) {
        long long p = s[i % k];
        p = (p * modPow(a, n - i)) % mod;
        p = (p * modPow(b, i)) % mod;
        res = (res + p) % mod;
      }
      break;
    }
  }
  res %= mod;
  res += mod;
  res %= mod;
  cout << res;
  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;
long long mod = (int)1e9 + 9;
long long modPow(long long x, long long k) {
  long long res = 1;
  for (x %= mod; k; k >>= 1) {
    if (k & 1) res = (res * x) % mod;
    x = (x * x) % mod;
  }
  return res;
}
long long modInv(long long x) { return modPow(x, mod - 2); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  k = min(k, n + 1);
  vector<long long> s(k);
  for (int i = 0; i < k; ++i) {
    char c;
    cin >> c;
    s[i] = c == '+' ? 1 : mod - 1;
  }
  while ((k << 1) < min(n, (long long)1e5)) {
    vector<long long> temp(s.begin(), s.end());
    s.insert(s.end(), temp.begin(), temp.end());
    k <<= 1;
  }
  long long dp = 0;
  for (int i = 0; i < k; ++i) {
    long long p = s[i];
    p = (p * modPow(a, n - i)) % mod;
    p = (p * modPow(b, i)) % mod;
    dp = (dp + p) % mod;
  }
  long long res = 0;
  for (long long i = 0; i <= n; i += k) {
    if (i + k <= n + 1) {
      res = (res + dp * modPow(b * modInv(a), i)) % mod;
    } else {
      for (; i <= n; ++i) {
        long long p = s[i % k];
        p = (p * modPow(a, n - i)) % mod;
        p = (p * modPow(b, i)) % mod;
        res = (res + p) % mod;
      }
      break;
    }
  }
  res %= mod;
  res += mod;
  res %= mod;
  cout << res;
  return 0;
}

```