#include <bits/stdc++.h>
using namespace std;
long long ara[100005];
string s;
long long bigmod(int a, int b) {
  if (b == 0) return 1;
  long long ret = bigmod(a, b / 2);
  ret = ret * ret;
  ret %= 1000000009;
  if (b % 2) ret *= a;
  ret %= 1000000009;
  return ret;
}
long long invmod(int a) { return bigmod(a, 1000000009 - 2); }
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '+')
      ara[i] = 1;
    else
      ara[i] = -1;
  }
  long long sum = 0;
  long long x = bigmod(b, k);
  long long y = bigmod(a, k);
  long long p = (x * invmod(y)) % 1000000009;
  long long d = (n + 1) / k;
  long long magic = 0;
  if (p % 1000000009 == 1) {
    magic = d;
  } else {
    magic = ((bigmod(p, d) - 1) * invmod(p - 1)) % 1000000009;
    magic = (magic + 1000000009) % 1000000009;
  }
  for (long long i = 0; i < k; i++) {
    long long val =
        (ara[i] * (bigmod(a, n - i) * bigmod(b, i)) % 1000000009) % 1000000009;
    val = (val + 1000000009) % 1000000009;
    val = (val * magic) % 1000000009;
    sum = sum + val;
    sum %= 1000000009;
  }
  sum = (sum + 1000000009) % 1000000009;
  cout << sum << endl;
  return 0;
}

### Prompt

Generate 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;
long long ara[100005];
string s;
long long bigmod(int a, int b) {
  if (b == 0) return 1;
  long long ret = bigmod(a, b / 2);
  ret = ret * ret;
  ret %= 1000000009;
  if (b % 2) ret *= a;
  ret %= 1000000009;
  return ret;
}
long long invmod(int a) { return bigmod(a, 1000000009 - 2); }
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '+')
      ara[i] = 1;
    else
      ara[i] = -1;
  }
  long long sum = 0;
  long long x = bigmod(b, k);
  long long y = bigmod(a, k);
  long long p = (x * invmod(y)) % 1000000009;
  long long d = (n + 1) / k;
  long long magic = 0;
  if (p % 1000000009 == 1) {
    magic = d;
  } else {
    magic = ((bigmod(p, d) - 1) * invmod(p - 1)) % 1000000009;
    magic = (magic + 1000000009) % 1000000009;
  }
  for (long long i = 0; i < k; i++) {
    long long val =
        (ara[i] * (bigmod(a, n - i) * bigmod(b, i)) % 1000000009) % 1000000009;
    val = (val + 1000000009) % 1000000009;
    val = (val * magic) % 1000000009;
    sum = sum + val;
    sum %= 1000000009;
  }
  sum = (sum + 1000000009) % 1000000009;
  cout << sum << endl;
  return 0;
}

```