#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 9;
const int N = 2e5 + 10;
long long powmod(long long a, long long b, long long mod = INF) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
long long powsum(long long a, long long b, long long mod = INF) {
  if (a == 1)
    return (b + 1) % mod;
  else
    return (powmod(a, b + 1) - 1) * powmod(a - 1, mod - 2) % mod;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string str;
  cin >> str;
  long long pre = powmod(a, n);
  a = powmod(a, INF - 2);
  a = a * b % INF;
  long long ans1 = 0;
  for (int i = 0; i < k; i++) {
    long long o = powmod(a, i);
    if (str[i] == '+')
      ans1 += o;
    else
      ans1 -= o;
    if (ans1 < 0) ans1 += INF;
    if (ans1 >= INF) ans1 -= INF;
  }
  int x = (n + 1) / k;
  long long o = powsum(powmod(a, k), x - 1);
  if (o < 0) o += INF;
  cout << o * pre % INF * ans1 % INF;
  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>
using namespace std;
const long long INF = 1e9 + 9;
const int N = 2e5 + 10;
long long powmod(long long a, long long b, long long mod = INF) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
long long powsum(long long a, long long b, long long mod = INF) {
  if (a == 1)
    return (b + 1) % mod;
  else
    return (powmod(a, b + 1) - 1) * powmod(a - 1, mod - 2) % mod;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string str;
  cin >> str;
  long long pre = powmod(a, n);
  a = powmod(a, INF - 2);
  a = a * b % INF;
  long long ans1 = 0;
  for (int i = 0; i < k; i++) {
    long long o = powmod(a, i);
    if (str[i] == '+')
      ans1 += o;
    else
      ans1 -= o;
    if (ans1 < 0) ans1 += INF;
    if (ans1 >= INF) ans1 -= INF;
  }
  int x = (n + 1) / k;
  long long o = powsum(powmod(a, k), x - 1);
  if (o < 0) o += INF;
  cout << o * pre % INF * ans1 % INF;
  return 0;
}

```