#include <bits/stdc++.h>
using namespace std;
const long long int mod = 998244353ll;
const long long int P = 1000000009ll;
const long long int N = 200005ll;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y) {
    if (y & 1) res = (res * x) % P;
    x = (x * x) % P;
    y /= 2;
  }
  return res;
}
void solve() {
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long int inv = (b * power(a, P - 2)) % P;
  long long int an = power(a, n);
  long long int ans = 0;
  long long int y = 1;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+')
      ans = (ans + an) % P;
    else
      ans = (ans + (P - an) % P) % P;
    y = (y * inv) % P;
    an = (an * inv) % P;
  }
  if (y == 1) {
    cout << (ans * ((n + 1) / k)) % P;
    return;
  }
  long long int d = (y - 1 + P) % P;
  d = power(d, P - 2);
  long long int num = (power(y, (n + 1) / k) % P - 1 + P) % P;
  num = (num * d) % P;
  num = (num * ans) % P;
  cout << num;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 int mod = 998244353ll;
const long long int P = 1000000009ll;
const long long int N = 200005ll;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y) {
    if (y & 1) res = (res * x) % P;
    x = (x * x) % P;
    y /= 2;
  }
  return res;
}
void solve() {
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long int inv = (b * power(a, P - 2)) % P;
  long long int an = power(a, n);
  long long int ans = 0;
  long long int y = 1;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+')
      ans = (ans + an) % P;
    else
      ans = (ans + (P - an) % P) % P;
    y = (y * inv) % P;
    an = (an * inv) % P;
  }
  if (y == 1) {
    cout << (ans * ((n + 1) / k)) % P;
    return;
  }
  long long int d = (y - 1 + P) % P;
  d = power(d, P - 2);
  long long int num = (power(y, (n + 1) / k) % P - 1 + P) % P;
  num = (num * d) % P;
  num = (num * ans) % P;
  cout << num;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}

```