#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000000 + 9;
long long int power(long long int a, long long int b, long long int mod) {
  a = a % mod;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      b--;
      res = (res * a) % mod;
    }
    a = a * a % mod;
    b = b >> 1;
  }
  return (res);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  long long int arr[100000];
  char temp;
  for (long long int i = 0; i < k; i++) {
    cin >> temp;
    if (temp == '+')
      arr[i] = 1;
    else
      arr[i] = -1;
  }
  long long int sum = 0;
  long long int x;
  for (long long int i = 0; i < k; i++) {
    x = (arr[i] * power(a, n - i, MOD) * power(b, i, MOD)) % MOD;
    if (x < 0) x = x + MOD;
    sum = (sum + x) % MOD;
  }
  long long int mul_sum;
  long long int z = (b % MOD * power(a, MOD - 2, MOD)) % MOD;
  z = power(z, k, MOD);
  if (z != 1) {
    mul_sum = ((long long int)power(z, (n + 1) / k, MOD) - 1) % MOD;
    if (mul_sum < 0) mul_sum = mul_sum + MOD;
    mul_sum = ((long long int)mul_sum * power(z - 1, MOD - 2, MOD)) % MOD;
  } else {
    mul_sum = (n + 1) / k;
  }
  sum = (sum * mul_sum) % MOD;
  if (sum >= 0)
    cout << sum;
  else
    cout << MOD + sum;
}

### Prompt

In CPP, your task is to solve 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 int MOD = 1000000000 + 9;
long long int power(long long int a, long long int b, long long int mod) {
  a = a % mod;
  long long int res = 1;
  while (b) {
    if (b & 1) {
      b--;
      res = (res * a) % mod;
    }
    a = a * a % mod;
    b = b >> 1;
  }
  return (res);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  long long int arr[100000];
  char temp;
  for (long long int i = 0; i < k; i++) {
    cin >> temp;
    if (temp == '+')
      arr[i] = 1;
    else
      arr[i] = -1;
  }
  long long int sum = 0;
  long long int x;
  for (long long int i = 0; i < k; i++) {
    x = (arr[i] * power(a, n - i, MOD) * power(b, i, MOD)) % MOD;
    if (x < 0) x = x + MOD;
    sum = (sum + x) % MOD;
  }
  long long int mul_sum;
  long long int z = (b % MOD * power(a, MOD - 2, MOD)) % MOD;
  z = power(z, k, MOD);
  if (z != 1) {
    mul_sum = ((long long int)power(z, (n + 1) / k, MOD) - 1) % MOD;
    if (mul_sum < 0) mul_sum = mul_sum + MOD;
    mul_sum = ((long long int)mul_sum * power(z - 1, MOD - 2, MOD)) % MOD;
  } else {
    mul_sum = (n + 1) / k;
  }
  sum = (sum * mul_sum) % MOD;
  if (sum >= 0)
    cout << sum;
  else
    cout << MOD + sum;
}

```