#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
const int mod = 1e9 + 9;
const int mxn = 1e5 + 9;
const long double eps = 1e-9;
long long qpow(long long n, long long k) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k >>= 1;
  }
  return ans % mod;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  long long i, j, m, n, k, a, b, p, ans = 1, sum = 0, t, x;
  string s;
  cin >> n >> a >> b >> k >> s;
  for (i = 0; i < k; i++) {
    t = qpow(a, n - i) * qpow(b, i) % mod;
    if (s[i] == '+')
      sum = (sum + t) % mod;
    else
      sum = (sum - t + mod) % mod;
  }
  x = (n + 1) / k - 1;
  t = (b % mod) * qpow(a, mod - 2) % mod;
  p = qpow(t, k) % mod;
  if (p == 1) {
    cout << sum * (x + 1) % mod << "\n";
    return 0;
  }
  ans = qpow(p, x + 1) % mod;
  ans = (ans - 1 + mod) % mod;
  ans = ans * qpow((p - 1 + mod) % mod, mod - 2) % mod;
  cout << (ans * sum % mod + mod) % mod << "\n";
  return 0;
}

### 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
const int mod = 1e9 + 9;
const int mxn = 1e5 + 9;
const long double eps = 1e-9;
long long qpow(long long n, long long k) {
  long long ans = 1;
  while (k > 0) {
    if (k & 1) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k >>= 1;
  }
  return ans % mod;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  long long i, j, m, n, k, a, b, p, ans = 1, sum = 0, t, x;
  string s;
  cin >> n >> a >> b >> k >> s;
  for (i = 0; i < k; i++) {
    t = qpow(a, n - i) * qpow(b, i) % mod;
    if (s[i] == '+')
      sum = (sum + t) % mod;
    else
      sum = (sum - t + mod) % mod;
  }
  x = (n + 1) / k - 1;
  t = (b % mod) * qpow(a, mod - 2) % mod;
  p = qpow(t, k) % mod;
  if (p == 1) {
    cout << sum * (x + 1) % mod << "\n";
    return 0;
  }
  ans = qpow(p, x + 1) % mod;
  ans = (ans - 1 + mod) % mod;
  ans = ans * qpow((p - 1 + mod) % mod, mod - 2) % mod;
  cout << (ans * sum % mod + mod) % mod << "\n";
  return 0;
}

```