#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 9;
long long binpow(long long x, long long y) {
  long long result = 1;
  while (y) {
    if (y % 2) result = (result * x) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return result;
}
long long inv(long long x) { return binpow(x, mod - 2); }
int main() {
  long long n, a, b, k, lim;
  vector<int> sign;
  char op;
  cin >> n >> a >> b >> k;
  long long sum = 0, acm = 0;
  sign.assign(k, 0);
  for (int i = 0; i < k; i++) {
    cin >> op;
    sign[i] = (((op == '+') ? 1 : -1));
  }
  lim = sqrt(n);
  lim = min(max(lim - lim % k, k), n + 1);
  for (int i = 0; i < lim; i++) {
    sum = ((sum + sign[i % k] * (binpow(a, n - i) * binpow(b, i)) % mod) % mod +
           mod) %
          mod;
  }
  acm = sum;
  if (lim < n) {
    for (int i = lim; i + lim <= n; i += lim) {
      acm =
          (acm + ((sum * binpow(b, i)) % mod * inv(binpow(a, i))) % mod) % mod;
    }
    for (int i = n - n % lim; i <= n; i++) {
      acm =
          ((acm + sign[i % k] * (binpow(a, n - i) * binpow(b, i)) % mod) % mod +
           mod) %
          mod;
    }
  }
  cout << (acm + mod) % mod << '\n';
  return 0;
}

### Prompt

Please create a solution in cpp 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 long long mod = 1e9 + 9;
long long binpow(long long x, long long y) {
  long long result = 1;
  while (y) {
    if (y % 2) result = (result * x) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return result;
}
long long inv(long long x) { return binpow(x, mod - 2); }
int main() {
  long long n, a, b, k, lim;
  vector<int> sign;
  char op;
  cin >> n >> a >> b >> k;
  long long sum = 0, acm = 0;
  sign.assign(k, 0);
  for (int i = 0; i < k; i++) {
    cin >> op;
    sign[i] = (((op == '+') ? 1 : -1));
  }
  lim = sqrt(n);
  lim = min(max(lim - lim % k, k), n + 1);
  for (int i = 0; i < lim; i++) {
    sum = ((sum + sign[i % k] * (binpow(a, n - i) * binpow(b, i)) % mod) % mod +
           mod) %
          mod;
  }
  acm = sum;
  if (lim < n) {
    for (int i = lim; i + lim <= n; i += lim) {
      acm =
          (acm + ((sum * binpow(b, i)) % mod * inv(binpow(a, i))) % mod) % mod;
    }
    for (int i = n - n % lim; i <= n; i++) {
      acm =
          ((acm + sign[i % k] * (binpow(a, n - i) * binpow(b, i)) % mod) % mod +
           mod) %
          mod;
    }
  }
  cout << (acm + mod) % mod << '\n';
  return 0;
}

```