#include <bits/stdc++.h>
using namespace std;
long long MD = 1000000009;
inline long long binpow(long long a, long long n) {
  long long res = 1LL;
  while (n) {
    if (n & 1LL) {
      res *= a;
      res %= MD;
    }
    a *= a;
    a %= MD;
    n >>= 1;
  }
  return res;
}
string s;
int n, k;
long long a, b;
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long sum = 0LL;
  long long aa = binpow(a, n);
  long long bb = 1LL;
  for (int i = 0; i < k; ++i) {
    long long cur = (aa * bb) % MD;
    if (s[i] == '+') {
      sum += cur;
      if (sum >= MD) sum -= MD;
    } else {
      sum -= cur;
      if (sum < 0) sum += MD;
    }
    bb *= b;
    bb %= MD;
    aa *= binpow(a, MD - 2);
    aa %= MD;
  }
  long long da = binpow(binpow(a, MD - 2), k);
  long long db = binpow(b, k);
  long long q = (da * db) % MD;
  if (q == 1LL) {
    long long res = sum * static_cast<long long>((n + 1) / k);
    res %= MD;
    cout << res << endl;
  } else {
    long long qn = binpow(q, (n + 1) / k);
    long long m1 = sum;
    long long m2 = (1 - qn + MD) % MD;
    long long m3 = binpow((1 - q + MD) % MD, MD - 2);
    long long res = (((m1 * m2) % MD) * m3) % MD;
    cout << res << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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 MD = 1000000009;
inline long long binpow(long long a, long long n) {
  long long res = 1LL;
  while (n) {
    if (n & 1LL) {
      res *= a;
      res %= MD;
    }
    a *= a;
    a %= MD;
    n >>= 1;
  }
  return res;
}
string s;
int n, k;
long long a, b;
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long sum = 0LL;
  long long aa = binpow(a, n);
  long long bb = 1LL;
  for (int i = 0; i < k; ++i) {
    long long cur = (aa * bb) % MD;
    if (s[i] == '+') {
      sum += cur;
      if (sum >= MD) sum -= MD;
    } else {
      sum -= cur;
      if (sum < 0) sum += MD;
    }
    bb *= b;
    bb %= MD;
    aa *= binpow(a, MD - 2);
    aa %= MD;
  }
  long long da = binpow(binpow(a, MD - 2), k);
  long long db = binpow(b, k);
  long long q = (da * db) % MD;
  if (q == 1LL) {
    long long res = sum * static_cast<long long>((n + 1) / k);
    res %= MD;
    cout << res << endl;
  } else {
    long long qn = binpow(q, (n + 1) / k);
    long long m1 = sum;
    long long m2 = (1 - qn + MD) % MD;
    long long m3 = binpow((1 - q + MD) % MD, MD - 2);
    long long res = (((m1 * m2) % MD) * m3) % MD;
    cout << res << endl;
  }
  return 0;
}

```