#include <bits/stdc++.h>
using namespace std;
const int inf = 2 * (1e9) + 10;
const int dx[] = {0, 0, -1, 1};
const int dy[] = {1, -1, 0, 0};
const long long MOD = 1000000009;
const int NMAX = 200005;
long long powbin(long long a, long long p) {
  long long first = 1;
  while (p) {
    if (p & 1) {
      first = (1LL * a * first) % MOD;
    }
    p >>= 1;
    a = (1LL * a * a) % MOD;
  }
  return (first + MOD) % MOD;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long pa = powbin(a, n);
  long long inva = powbin(a, MOD - 2);
  long long prev = 0;
  long long pb = 1;
  char ch;
  for (int i = 0; i < k; ++i) {
    cin >> ch;
    if (ch == '+') {
      prev = (MOD + prev + (1LL * pa * pb) % MOD) % MOD;
    } else {
      prev = (MOD + prev - (1LL * pa * pb) % MOD) % MOD;
    }
    pa = (1LL * inva * pa) % MOD;
    pb = (1LL * b * pb) % MOD;
    prev = (prev + MOD) % MOD;
  }
  long long m = (n + 1) / k;
  long long num = (MOD + 1 - powbin((1LL * inva * b) % MOD, k * m)) % MOD;
  long long den = (MOD + 1 - powbin((1LL * inva * b) % MOD, k)) % MOD;
  num = (num + MOD) % MOD;
  den = (den + MOD) % MOD;
  long long quo = (1LL * num * powbin(den, MOD - 2)) % MOD;
  quo = (quo + MOD) % MOD;
  if (den == 0) {
    cout << (1LL * prev * m) % MOD;
  } else
    cout << ((1LL * prev * quo) % MOD + MOD) % MOD << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 int inf = 2 * (1e9) + 10;
const int dx[] = {0, 0, -1, 1};
const int dy[] = {1, -1, 0, 0};
const long long MOD = 1000000009;
const int NMAX = 200005;
long long powbin(long long a, long long p) {
  long long first = 1;
  while (p) {
    if (p & 1) {
      first = (1LL * a * first) % MOD;
    }
    p >>= 1;
    a = (1LL * a * a) % MOD;
  }
  return (first + MOD) % MOD;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long pa = powbin(a, n);
  long long inva = powbin(a, MOD - 2);
  long long prev = 0;
  long long pb = 1;
  char ch;
  for (int i = 0; i < k; ++i) {
    cin >> ch;
    if (ch == '+') {
      prev = (MOD + prev + (1LL * pa * pb) % MOD) % MOD;
    } else {
      prev = (MOD + prev - (1LL * pa * pb) % MOD) % MOD;
    }
    pa = (1LL * inva * pa) % MOD;
    pb = (1LL * b * pb) % MOD;
    prev = (prev + MOD) % MOD;
  }
  long long m = (n + 1) / k;
  long long num = (MOD + 1 - powbin((1LL * inva * b) % MOD, k * m)) % MOD;
  long long den = (MOD + 1 - powbin((1LL * inva * b) % MOD, k)) % MOD;
  num = (num + MOD) % MOD;
  den = (den + MOD) % MOD;
  long long quo = (1LL * num * powbin(den, MOD - 2)) % MOD;
  quo = (quo + MOD) % MOD;
  if (den == 0) {
    cout << (1LL * prev * m) % MOD;
  } else
    cout << ((1LL * prev * quo) % MOD + MOD) % MOD << "\n";
  return 0;
}

```