#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 9;
int n, a, b, k, s[N], ans, last;
inline int mul(int x, int y) { return (long long int)x * y % mod; }
int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
int fast_pow(int x, int y) {
  if (y == 0) return 1;
  int res = fast_pow(x, y / 2);
  res = mul(res, res);
  if (y & 1) res = mul(res, x);
  return res;
}
int inv(int x) { return fast_pow(x, mod - 2); }
int get(int x, int y) {
  if (x == 1) return y;
  return mul(add(fast_pow(x, y), mod - 1), inv(add(x, mod - 1)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> a >> b >> k;
  for (int i = 0; i < k; i++) {
    char ch;
    cin >> ch;
    s[i] = (ch == '-' ? mod - 1 : 1);
  }
  for (int i = 0; i < k; i++)
    last = add(last, mul(s[i], mul(fast_pow(b, i), fast_pow(a, n - i))));
  int m = mul(fast_pow(b, k), inv(fast_pow(a, k)));
  cout << mul(last, get(m, (n + 1) / k)) << endl;
}

### 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 int N = 1e5 + 5;
const int mod = 1e9 + 9;
int n, a, b, k, s[N], ans, last;
inline int mul(int x, int y) { return (long long int)x * y % mod; }
int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
int fast_pow(int x, int y) {
  if (y == 0) return 1;
  int res = fast_pow(x, y / 2);
  res = mul(res, res);
  if (y & 1) res = mul(res, x);
  return res;
}
int inv(int x) { return fast_pow(x, mod - 2); }
int get(int x, int y) {
  if (x == 1) return y;
  return mul(add(fast_pow(x, y), mod - 1), inv(add(x, mod - 1)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> a >> b >> k;
  for (int i = 0; i < k; i++) {
    char ch;
    cin >> ch;
    s[i] = (ch == '-' ? mod - 1 : 1);
  }
  for (int i = 0; i < k; i++)
    last = add(last, mul(s[i], mul(fast_pow(b, i), fast_pow(a, n - i))));
  int m = mul(fast_pow(b, k), inv(fast_pow(a, k)));
  cout << mul(last, get(m, (n + 1) / k)) << endl;
}

```