#include <bits/stdc++.h>
using namespace std;
constexpr int MAX = 200010;
constexpr int MOD = 1000000009;
long long n, a, b, k;
string s;
long long modpow(long long a, long long b) {
  long long res = 1, p = a;
  while (b > 0) {
    if (b & 1) {
      res = (res * p) % MOD;
    }
    b >>= 1;
    p = (p * p) % MOD;
  }
  return res;
}
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long a_inv = modpow(a, MOD - 2);
  long long z = 0;
  long long t1 = modpow(a, n), t2 = 1;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+') {
      z = (z + t1 * t2) % MOD;
    } else {
      z = ((z - t1 * t2) % MOD + MOD) % MOD;
    }
    t1 = (t1 * a_inv) % MOD;
    t2 = (t2 * b) % MOD;
  }
  long long q = modpow((a_inv * b) % MOD, k);
  long long res = 0;
  if (q != 1) {
    res = (modpow(q, (n + 1) / k) - 1 + MOD) % MOD;
    res = (res * modpow(q - 1, MOD - 2)) % MOD;
  } else {
    res = (n + 1) / k;
  }
  res = (res * z) % MOD;
  cout << res << endl;
  return 0;
}

### Prompt

Your task is to create 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;
constexpr int MAX = 200010;
constexpr int MOD = 1000000009;
long long n, a, b, k;
string s;
long long modpow(long long a, long long b) {
  long long res = 1, p = a;
  while (b > 0) {
    if (b & 1) {
      res = (res * p) % MOD;
    }
    b >>= 1;
    p = (p * p) % MOD;
  }
  return res;
}
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long a_inv = modpow(a, MOD - 2);
  long long z = 0;
  long long t1 = modpow(a, n), t2 = 1;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+') {
      z = (z + t1 * t2) % MOD;
    } else {
      z = ((z - t1 * t2) % MOD + MOD) % MOD;
    }
    t1 = (t1 * a_inv) % MOD;
    t2 = (t2 * b) % MOD;
  }
  long long q = modpow((a_inv * b) % MOD, k);
  long long res = 0;
  if (q != 1) {
    res = (modpow(q, (n + 1) / k) - 1 + MOD) % MOD;
    res = (res * modpow(q - 1, MOD - 2)) % MOD;
  } else {
    res = (n + 1) / k;
  }
  res = (res * z) % MOD;
  cout << res << endl;
  return 0;
}

```