#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000009;
long long md(long long n) {
  if (n >= mod) return n % mod;
  return n;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, k, inv;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  inv = power(a, mod - 2);
  long long term = power(a, n), sum = 0;
  for (long long i = 0; s[i]; i++) {
    if (s[i] == '+')
      sum = md(sum + term);
    else
      sum = md(sum - term + mod);
    term = md(md(term * inv) * b);
  }
  long long ans;
  long long A = sum, N = (n + 1) / k, R = power(md(b * inv), k);
  if (R == 1) {
    ans = sum * (n + 1) / k;
    ans %= mod;
  } else {
    long long numerator = md(A * md((power(R, N) - 1 + mod)));
    long long den = power(R - 1, mod - 2);
    numerator = md(numerator * den);
    ans = numerator;
  }
  cout << ans;
}

### 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;
const long long mod = 1000000009;
long long md(long long n) {
  if (n >= mod) return n % mod;
  return n;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, k, inv;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  inv = power(a, mod - 2);
  long long term = power(a, n), sum = 0;
  for (long long i = 0; s[i]; i++) {
    if (s[i] == '+')
      sum = md(sum + term);
    else
      sum = md(sum - term + mod);
    term = md(md(term * inv) * b);
  }
  long long ans;
  long long A = sum, N = (n + 1) / k, R = power(md(b * inv), k);
  if (R == 1) {
    ans = sum * (n + 1) / k;
    ans %= mod;
  } else {
    long long numerator = md(A * md((power(R, N) - 1 + mod)));
    long long den = power(R - 1, mod - 2);
    numerator = md(numerator * den);
    ans = numerator;
  }
  cout << ans;
}

```