#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000009;
char c[200000];
long long n, a, b, k, ans;
long long mypow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2) return (mypow(a, n - 1) * a) % mod;
  long long res = mypow(a, n / 2);
  return (res * res) % mod;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cin >> n >> a >> b >> k;
  n += 1;
  long long a1 = mypow(b, k);
  long long b1 = mypow(mypow(a, k), mod - 2);
  long long q = (a1 * b1) % mod;
  vector<char> c(k + 1);
  for (long long i = 0; i < k; i++) {
    cin >> c[i];
    long long a2 = mypow(a, n - 1 - i);
    long long b2 = mypow(b, i);
    long long base = (a2 * b2) % mod;
    if (q != 1) {
      long long ans1 = 1 + mod - mypow(q, n / k);
      ans1 %= mod;
      long long ans2 = mypow((1 - q + mod) % mod, mod - 2);
      ans += ((c[i] == '-') ? -1 : 1) * ((base * ((ans1 * ans2) % mod)) % mod);
    } else {
      ans += ((c[i] == '-') ? -1 : 1) * (((n / k) * base) % mod);
    }
    ans = (ans + mod * 2) % mod;
  }
  cout << ans;
  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;
long long mod = 1000000009;
char c[200000];
long long n, a, b, k, ans;
long long mypow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2) return (mypow(a, n - 1) * a) % mod;
  long long res = mypow(a, n / 2);
  return (res * res) % mod;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cin >> n >> a >> b >> k;
  n += 1;
  long long a1 = mypow(b, k);
  long long b1 = mypow(mypow(a, k), mod - 2);
  long long q = (a1 * b1) % mod;
  vector<char> c(k + 1);
  for (long long i = 0; i < k; i++) {
    cin >> c[i];
    long long a2 = mypow(a, n - 1 - i);
    long long b2 = mypow(b, i);
    long long base = (a2 * b2) % mod;
    if (q != 1) {
      long long ans1 = 1 + mod - mypow(q, n / k);
      ans1 %= mod;
      long long ans2 = mypow((1 - q + mod) % mod, mod - 2);
      ans += ((c[i] == '-') ? -1 : 1) * ((base * ((ans1 * ans2) % mod)) % mod);
    } else {
      ans += ((c[i] == '-') ? -1 : 1) * (((n / k) * base) % mod);
    }
    ans = (ans + mod * 2) % mod;
  }
  cout << ans;
  return 0;
}

```