#include <bits/stdc++.h>
using namespace std;
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long m = 1e9 + 9;
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long ans = 0;
  string s;
  cin >> s;
  for (int i = 0; i < k; i++) {
    long long p = binpow(a, n - i, m);
    p = (p % m * binpow(b, i, m));
    if (s[i] == '+')
      ans = (ans + p) % m;
    else
      ans = (ans - p + m) % m;
  }
  long long q = (n + 1) / k;
  long long r = binpow(b, k, m);
  r = (r % m * binpow(binpow(a, k, m), m - 2, m) % m) % m;
  if (r == 1) {
    long long aa = ans;
    aa = (aa % m * q % m) % m;
    cout << (aa + m) % m << "\n";
    return 0;
  }
  ans = (ans % m * (binpow(r, q, m) - 1 + m) % m) % m;
  ans = (ans % m * binpow((r - 1 + m) % m, m - 2, m)) % m;
  cout << (ans + m) % m << "\n";
}

### 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;
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long m = 1e9 + 9;
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long ans = 0;
  string s;
  cin >> s;
  for (int i = 0; i < k; i++) {
    long long p = binpow(a, n - i, m);
    p = (p % m * binpow(b, i, m));
    if (s[i] == '+')
      ans = (ans + p) % m;
    else
      ans = (ans - p + m) % m;
  }
  long long q = (n + 1) / k;
  long long r = binpow(b, k, m);
  r = (r % m * binpow(binpow(a, k, m), m - 2, m) % m) % m;
  if (r == 1) {
    long long aa = ans;
    aa = (aa % m * q % m) % m;
    cout << (aa + m) % m << "\n";
    return 0;
  }
  ans = (ans % m * (binpow(r, q, m) - 1 + m) % m) % m;
  ans = (ans % m * binpow((r - 1 + m) % m, m - 2, m)) % m;
  cout << (ans + m) % m << "\n";
}

```