#include <bits/stdc++.h>
using namespace std;
long long mexp(long long a, long long p) {
  if (p == 0) return 1;
  long long ans = mexp(a, p / 2);
  ans = (ans * ans) % 1000000009;
  if (p % 2)
    return (a * ans) % 1000000009;
  else
    return ans;
}
void solve() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string second;
  cin >> second;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    int m = 1;
    if (second[i] == '-') m = -1;
    ans += m * (mexp(a, n - i) * mexp(b, i)) % 1000000009;
    if (ans < 0) ans += 1000000009;
    ans = ans % 1000000009;
  }
  long long pp = (n + 1) / k;
  long long first = (mexp(a, 1000000009 - 1 - k) * mexp(b, k)) % 1000000009;
  if (first == 1) {
    cout << (pp * ans) % 1000000009;
    return;
  }
  long long ff = (mexp(first, pp) - 1);
  if (ff < 0) ff += 1000000009;
  first = (ff * mexp(first - 1, 1000000009 - 2)) % 1000000009;
  cout << (first * ans) % 1000000009;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(12);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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 mexp(long long a, long long p) {
  if (p == 0) return 1;
  long long ans = mexp(a, p / 2);
  ans = (ans * ans) % 1000000009;
  if (p % 2)
    return (a * ans) % 1000000009;
  else
    return ans;
}
void solve() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string second;
  cin >> second;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    int m = 1;
    if (second[i] == '-') m = -1;
    ans += m * (mexp(a, n - i) * mexp(b, i)) % 1000000009;
    if (ans < 0) ans += 1000000009;
    ans = ans % 1000000009;
  }
  long long pp = (n + 1) / k;
  long long first = (mexp(a, 1000000009 - 1 - k) * mexp(b, k)) % 1000000009;
  if (first == 1) {
    cout << (pp * ans) % 1000000009;
    return;
  }
  long long ff = (mexp(first, pp) - 1);
  if (ff < 0) ff += 1000000009;
  first = (ff * mexp(first - 1, 1000000009 - 2)) % 1000000009;
  cout << (first * ans) % 1000000009;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(12);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}

```