#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 9;
const int maxn = 1e5 + 7;
char s[maxn];
int quickPow(int a, int times) {
  int rt = 1, base = a;
  while (times) {
    if (times & 1) rt = (long long)rt * base % M;
    base = (long long)base * base % M;
    times >>= 1;
  }
  return rt;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  long long tmp = (n + 1) / k;
  long long inv_a = quickPow(a, M - 2);
  long long head = quickPow(a, n);
  long long gap = (long long)quickPow(b, k) * quickPow(inv_a, k) % M;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    int flag = 1;
    if (s[i] == '-') flag = -1;
    if (gap == 1)
      ans = (ans + flag * (long long)head * tmp) % M;
    else
      ans = (ans + flag * (long long)head * (1 - quickPow(gap, tmp)) % M *
                       quickPow(1 - gap, M - 2) % M) %
            M;
    head = (long long)head * b % M * inv_a % M;
  }
  cout << (ans + M) % M << 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;
const int M = 1e9 + 9;
const int maxn = 1e5 + 7;
char s[maxn];
int quickPow(int a, int times) {
  int rt = 1, base = a;
  while (times) {
    if (times & 1) rt = (long long)rt * base % M;
    base = (long long)base * base % M;
    times >>= 1;
  }
  return rt;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  long long tmp = (n + 1) / k;
  long long inv_a = quickPow(a, M - 2);
  long long head = quickPow(a, n);
  long long gap = (long long)quickPow(b, k) * quickPow(inv_a, k) % M;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    int flag = 1;
    if (s[i] == '-') flag = -1;
    if (gap == 1)
      ans = (ans + flag * (long long)head * tmp) % M;
    else
      ans = (ans + flag * (long long)head * (1 - quickPow(gap, tmp)) % M *
                       quickPow(1 - gap, M - 2) % M) %
            M;
    head = (long long)head * b % M * inv_a % M;
  }
  cout << (ans + M) % M << endl;
  return 0;
}

```