#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000009;
int binpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ret;
}
int calc(int a, int b, int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    return (1LL * b * calc(a, b, n - 1) + binpow(a, n)) % mod;
  }
  return 1LL * (a + b) * calc(1LL * a * a % mod, 1LL * b * b % mod, n / 2) %
         mod;
}
int main() {
  int n, a, b, k;
  string s;
  cin >> n >> a >> b >> k >> s;
  swap(a, b);
  int ra = a, rb = b;
  a = binpow(a, k);
  b = binpow(b, k);
  int ret = 0;
  for (int i = 0; i < k; ++i) {
    int r = n / k + 1;
    if (1LL * r * k + i > n) --r;
    int res = calc(a, b, r);
    int ea = i, eb = n % k - i;
    if (eb < 0) eb += k;
    int aug = 1LL * res * binpow(ra, ea) % mod * binpow(rb, eb) % mod;
    if (s[i] == '+')
      ret = (ret + aug) % mod;
    else
      ret = (ret + mod - aug) % mod;
  }
  cout << ret << '\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 mod = 1000000009;
int binpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ret;
}
int calc(int a, int b, int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    return (1LL * b * calc(a, b, n - 1) + binpow(a, n)) % mod;
  }
  return 1LL * (a + b) * calc(1LL * a * a % mod, 1LL * b * b % mod, n / 2) %
         mod;
}
int main() {
  int n, a, b, k;
  string s;
  cin >> n >> a >> b >> k >> s;
  swap(a, b);
  int ra = a, rb = b;
  a = binpow(a, k);
  b = binpow(b, k);
  int ret = 0;
  for (int i = 0; i < k; ++i) {
    int r = n / k + 1;
    if (1LL * r * k + i > n) --r;
    int res = calc(a, b, r);
    int ea = i, eb = n % k - i;
    if (eb < 0) eb += k;
    int aug = 1LL * res * binpow(ra, ea) % mod * binpow(rb, eb) % mod;
    if (s[i] == '+')
      ret = (ret + aug) % mod;
    else
      ret = (ret + mod - aug) % mod;
  }
  cout << ret << '\n';
  return 0;
}

```