#include <bits/stdc++.h>
using namespace std;
long long num, r, n, a, b, k, mod = 1000000009;
long long fastp(long long base, long long exp) {
  if (exp == 0) return 1ll;
  long long tmp = fastp(base, exp / 2ll);
  tmp = (tmp * tmp) % mod;
  if (exp % 2 == 1) tmp *= base;
  return tmp % mod;
}
long long inv(long long base) { return fastp(base, mod - 2ll); }
int main() {
  cin >> n >> a >> b >> k;
  num = (n + 1ll) / k;
  r = fastp(b, k) * inv(fastp(a, k));
  r %= mod;
  char c;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    cin >> c;
    long long val = (((fastp(r, num) - 1ll) * inv(r - 1ll)) % mod);
    if (r == 1) {
      if (c == '+')
        ans += (num * ((fastp(a, n - i) * fastp(b, i)) % mod)) % mod;
      else
        ans -= (num * ((fastp(a, n - i) * fastp(b, i)) % mod)) % mod;
    } else {
      if (c == '+')
        ans +=
            ((val * ((fastp(a, n - i) * fastp(b, i)) % mod)) % mod + mod) % mod;
      else
        ans -=
            ((val * ((fastp(a, n - i) * fastp(b, i)) % mod)) % mod + mod) % mod;
    }
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 num, r, n, a, b, k, mod = 1000000009;
long long fastp(long long base, long long exp) {
  if (exp == 0) return 1ll;
  long long tmp = fastp(base, exp / 2ll);
  tmp = (tmp * tmp) % mod;
  if (exp % 2 == 1) tmp *= base;
  return tmp % mod;
}
long long inv(long long base) { return fastp(base, mod - 2ll); }
int main() {
  cin >> n >> a >> b >> k;
  num = (n + 1ll) / k;
  r = fastp(b, k) * inv(fastp(a, k));
  r %= mod;
  char c;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    cin >> c;
    long long val = (((fastp(r, num) - 1ll) * inv(r - 1ll)) % mod);
    if (r == 1) {
      if (c == '+')
        ans += (num * ((fastp(a, n - i) * fastp(b, i)) % mod)) % mod;
      else
        ans -= (num * ((fastp(a, n - i) * fastp(b, i)) % mod)) % mod;
    } else {
      if (c == '+')
        ans +=
            ((val * ((fastp(a, n - i) * fastp(b, i)) % mod)) % mod + mod) % mod;
      else
        ans -=
            ((val * ((fastp(a, n - i) * fastp(b, i)) % mod)) % mod + mod) % mod;
    }
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}

```