#include <bits/stdc++.h>
using namespace std;
long long p = 1000000000 + 9;
long long binpow(long long a, long long n) {
  if (n == 1) {
    return (a % p + p) % p;
  }
  if (n % 2 == 0) {
    long long b = binpow(a, n / 2);
    return ((b * b) % p + p) % p;
  } else {
    long long b = binpow(a, n - 1);
    return ((a * b) % p + p) % p;
  }
}
void read(vector<long long> &a) {
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
  }
}
void write(vector<long long> &a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << ' ';
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long t = 0;
  long long x = binpow(a, n);
  long long z = binpow(a, p - 2);
  long long y = 1;
  for (int i = 0; i < k; i++) {
    char c;
    cin >> c;
    long long kek;
    if (c == '+') {
      kek = 1;
    } else {
      kek = p - 1;
    }
    t = ((t + (((kek * x) % p) * y) % p + p) % p + p) % p;
    x = ((x * z) % p + p) % p;
    y = ((y * b) % p + p) % p;
  }
  long long drob = (binpow((b * z) % p, k) + p) % p;
  if (drob == 1) {
    cout << ((t * (n + 1) / k) % p + p) % p;
    return 0;
  }
  long long chisl = ((binpow(drob, (n + 1) / k) - 1 + p) % p + p) % p;
  long long znam = ((drob - 1 + p) % p + p) % p;
  cout << ((((t * chisl) % p) * binpow(znam, p - 2)) % p + p) % p;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 p = 1000000000 + 9;
long long binpow(long long a, long long n) {
  if (n == 1) {
    return (a % p + p) % p;
  }
  if (n % 2 == 0) {
    long long b = binpow(a, n / 2);
    return ((b * b) % p + p) % p;
  } else {
    long long b = binpow(a, n - 1);
    return ((a * b) % p + p) % p;
  }
}
void read(vector<long long> &a) {
  for (int i = 0; i < a.size(); i++) {
    cin >> a[i];
  }
}
void write(vector<long long> &a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << ' ';
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long t = 0;
  long long x = binpow(a, n);
  long long z = binpow(a, p - 2);
  long long y = 1;
  for (int i = 0; i < k; i++) {
    char c;
    cin >> c;
    long long kek;
    if (c == '+') {
      kek = 1;
    } else {
      kek = p - 1;
    }
    t = ((t + (((kek * x) % p) * y) % p + p) % p + p) % p;
    x = ((x * z) % p + p) % p;
    y = ((y * b) % p + p) % p;
  }
  long long drob = (binpow((b * z) % p, k) + p) % p;
  if (drob == 1) {
    cout << ((t * (n + 1) / k) % p + p) % p;
    return 0;
  }
  long long chisl = ((binpow(drob, (n + 1) / k) - 1 + p) % p + p) % p;
  long long znam = ((drob - 1 + p) % p + p) % p;
  cout << ((((t * chisl) % p) * binpow(znam, p - 2)) % p + p) % p;
}

```