#include <bits/stdc++.h>
using namespace std;
long long modpow(long long a, long long b,
                 long long mod = (long long)(1e9 + 7)) {
  if (!b) return 1;
  a %= mod;
  return modpow(a * a % mod, b / 2, mod) * (b & 1 ? a : 1) % mod;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long mod = 1e9 + 9;
long long n, a, b, k;
string s;
long long inv(long long a) { return modpow(a, mod - 2, mod); }
void solve() {
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long y = inv(a) * b % mod;
    if (modpow(y, k, mod) != 1) {
      ans = (ans + mod +
             (s[i] == '+' ? 1 : -1) * modpow(a, n, mod) *
                 (modpow(y, n + 1, mod) + mod - 1) % mod *
                 inv(modpow(y, k, mod) + mod - 1) % mod * modpow(y, i, mod) %
                 mod) %
            mod;
    } else {
      ans = (ans + mod +
             (s[i] == '+' ? 1 : -1) * modpow(a, n, mod) * ((n + 1) / k) % mod *
                 modpow(y, i, mod) % mod) %
            mod;
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  while (cin >> n >> a >> b >> k >> s) solve();
  return 0;
}

### Prompt

Your challenge is to write 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;
long long modpow(long long a, long long b,
                 long long mod = (long long)(1e9 + 7)) {
  if (!b) return 1;
  a %= mod;
  return modpow(a * a % mod, b / 2, mod) * (b & 1 ? a : 1) % mod;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long mod = 1e9 + 9;
long long n, a, b, k;
string s;
long long inv(long long a) { return modpow(a, mod - 2, mod); }
void solve() {
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long y = inv(a) * b % mod;
    if (modpow(y, k, mod) != 1) {
      ans = (ans + mod +
             (s[i] == '+' ? 1 : -1) * modpow(a, n, mod) *
                 (modpow(y, n + 1, mod) + mod - 1) % mod *
                 inv(modpow(y, k, mod) + mod - 1) % mod * modpow(y, i, mod) %
                 mod) %
            mod;
    } else {
      ans = (ans + mod +
             (s[i] == '+' ? 1 : -1) * modpow(a, n, mod) * ((n + 1) / k) % mod *
                 modpow(y, i, mod) % mod) %
            mod;
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  while (cin >> n >> a >> b >> k >> s) solve();
  return 0;
}

```