#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
const int mod = 1e9 + 9;
long long int pow_mod(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) {
      ans = 1LL * ans * a % mod;
      ans = ans % mod;
    }
    b = b >> 1;
    a = a * a % mod;
  }
  return ans;
}
long long int n, a, b, k;
string s;
int main(void) {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  cin >> s;
  long long int seg_value = 0;
  for (int i = 0; i < k; ++i) {
    (seg_value += ((s[i] == '+' ? 1 : -1)) * pow_mod(a, n - i) % mod *
                  pow_mod(b, i) % mod) %= mod;
  }
  long long int z = pow_mod(b, k) * pow_mod(pow_mod(a, mod - 2), k) % mod;
  z = (z + mod) % mod;
  if (z == 1)
    seg_value = seg_value * ((n + 1) / k) % mod;
  else {
    seg_value =
        seg_value *
        ((pow_mod(z, (n + 1) / k) - 1) * pow_mod(z - 1, mod - 2) % mod) % mod;
  }
  seg_value = (seg_value + mod) % mod;
  cout << seg_value << endl;
}

### Prompt

Your challenge is to write 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 int N = 5e5 + 5;
const int mod = 1e9 + 9;
long long int pow_mod(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) {
      ans = 1LL * ans * a % mod;
      ans = ans % mod;
    }
    b = b >> 1;
    a = a * a % mod;
  }
  return ans;
}
long long int n, a, b, k;
string s;
int main(void) {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  cin >> s;
  long long int seg_value = 0;
  for (int i = 0; i < k; ++i) {
    (seg_value += ((s[i] == '+' ? 1 : -1)) * pow_mod(a, n - i) % mod *
                  pow_mod(b, i) % mod) %= mod;
  }
  long long int z = pow_mod(b, k) * pow_mod(pow_mod(a, mod - 2), k) % mod;
  z = (z + mod) % mod;
  if (z == 1)
    seg_value = seg_value * ((n + 1) / k) % mod;
  else {
    seg_value =
        seg_value *
        ((pow_mod(z, (n + 1) / k) - 1) * pow_mod(z - 1, mod - 2) % mod) % mod;
  }
  seg_value = (seg_value + mod) % mod;
  cout << seg_value << endl;
}

```