#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
long long power(long long x, long long y) {
  if (x == 0) return 0;
  long long ret = 1;
  for (; y; y >>= 1, x = x * x % 1000000009)
    if (y & 1) (ret *= x) %= 1000000009;
  return ret;
}
int main() {
  long long n, a, b, k;
  while (cin >> n >> a >> b >> k) {
    string ch;
    cin >> ch;
    int len = ch.length();
    long long aa = power(a, n), tmp = 0, bb = 1LL;
    long long ret_a = power(a, k), ret_b = power(b, k);
    for (int i = 0; i < len; i++) {
      if (ch[i] == '+')
        tmp = (tmp + aa * bb % 1000000009) % 1000000009;
      else
        tmp = (tmp - aa * bb % 1000000009 + 1000000009) % 1000000009;
      (aa *= power(a, 1000000009 - 2)) %= 1000000009;
      (bb *= b) %= 1000000009;
    }
    long long q = ret_b * power(ret_a, 1000000009 - 2) % 1000000009;
    long long sz = (n + 1) / k;
    long long ans = 0;
    if (q != 1) {
      ans = (power(q, sz) - 1 + 1000000009) % 1000000009 *
            power(q - 1, 1000000009 - 2) % 1000000009;
      (ans *= tmp) %= 1000000009;
      cout << ans << endl;
    } else {
      ans = (tmp * sz) % 1000000009;
      cout << ans << endl;
    }
  }
  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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
long long power(long long x, long long y) {
  if (x == 0) return 0;
  long long ret = 1;
  for (; y; y >>= 1, x = x * x % 1000000009)
    if (y & 1) (ret *= x) %= 1000000009;
  return ret;
}
int main() {
  long long n, a, b, k;
  while (cin >> n >> a >> b >> k) {
    string ch;
    cin >> ch;
    int len = ch.length();
    long long aa = power(a, n), tmp = 0, bb = 1LL;
    long long ret_a = power(a, k), ret_b = power(b, k);
    for (int i = 0; i < len; i++) {
      if (ch[i] == '+')
        tmp = (tmp + aa * bb % 1000000009) % 1000000009;
      else
        tmp = (tmp - aa * bb % 1000000009 + 1000000009) % 1000000009;
      (aa *= power(a, 1000000009 - 2)) %= 1000000009;
      (bb *= b) %= 1000000009;
    }
    long long q = ret_b * power(ret_a, 1000000009 - 2) % 1000000009;
    long long sz = (n + 1) / k;
    long long ans = 0;
    if (q != 1) {
      ans = (power(q, sz) - 1 + 1000000009) % 1000000009 *
            power(q - 1, 1000000009 - 2) % 1000000009;
      (ans *= tmp) %= 1000000009;
      cout << ans << endl;
    } else {
      ans = (tmp * sz) % 1000000009;
      cout << ans << endl;
    }
  }
  return 0;
}

```