#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 9;
long long poww(long long a, long long b, long long c) {
  long long ans = 1 % c;
  for (a %= c; b; b >>= 1, a = a * a % c)
    if (b & 1) ans = ans * a % c;
  return ans;
}
int main() {
  long long n, a, b, k;
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  string s;
  cin >> s;
  long long sum = 0;
  long long t;
  for (int i = 0; i < s.size(); i++) {
    t = poww(a, n - i, mod) * poww(b, i, mod);
    if (s[i] == '+')
      sum = (sum + t) % mod;
    else
      sum = (sum - t + mod) % mod;
  }
  long long num = (n + 1) / k;
  long long aa = poww(a, k, mod), bb = poww(b, k, mod);
  aa = poww(aa, mod - 2, mod);
  long long q = bb * aa % mod;
  if (q == 1) {
    printf("%lld\n", (sum * num % mod + mod) % mod);
    return 0;
  }
  long long qq = poww(q - 1, mod - 2, mod);
  long long qqq = (poww(q, num, mod) - 1) * sum % mod;
  printf("%lld\n", (qq * qqq % mod + mod) % mod);
}

### Prompt

Develop a solution in cpp 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 = 1e9 + 9;
long long poww(long long a, long long b, long long c) {
  long long ans = 1 % c;
  for (a %= c; b; b >>= 1, a = a * a % c)
    if (b & 1) ans = ans * a % c;
  return ans;
}
int main() {
  long long n, a, b, k;
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  string s;
  cin >> s;
  long long sum = 0;
  long long t;
  for (int i = 0; i < s.size(); i++) {
    t = poww(a, n - i, mod) * poww(b, i, mod);
    if (s[i] == '+')
      sum = (sum + t) % mod;
    else
      sum = (sum - t + mod) % mod;
  }
  long long num = (n + 1) / k;
  long long aa = poww(a, k, mod), bb = poww(b, k, mod);
  aa = poww(aa, mod - 2, mod);
  long long q = bb * aa % mod;
  if (q == 1) {
    printf("%lld\n", (sum * num % mod + mod) % mod);
    return 0;
  }
  long long qq = poww(q - 1, mod - 2, mod);
  long long qqq = (poww(q, num, mod) - 1) * sum % mod;
  printf("%lld\n", (qq * qqq % mod + mod) % mod);
}

```