#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
constexpr auto MOD = 1000000009LL;
constexpr auto LINF = (1LL << 60);
constexpr auto INF = 2147483647LL;
constexpr auto PI = 3.1415926535897932384626433;
constexpr auto EPS = 1E-9;
template <typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2> p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
auto printVector = []<typename T>(ostream& out, vector<T> v) {
  copy(v.begin(), v.end(), ostream_iterator<T>(out, " "));
};
ll pwr(ll b, ll e) {
  b %= MOD;
  if (e == 0) return 1;
  if (e == 1) return b;
  return (pwr(b * b, e / 2) * pwr(b, e % 2)) % MOD;
}
int n, a, b, k;
string str;
ll res, mve, kmove, now;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> a >> b >> k;
  string str;
  cin >> str;
  mve = pwr(a, MOD - 2) * b % MOD;
  kmove = pwr(mve, k);
  now = pwr(a, n);
  now = (kmove == 1 ? now * (n + 1) / k % MOD
                    : now * (1 - pwr(kmove, (n + 1) / k)) % MOD *
                          pwr(1 - kmove, MOD - 2) % MOD);
  for (int i = 0; i < k; i++) {
    (res += (str[i] == '+' ? now : -now)) %= MOD;
    (now *= mve) %= MOD;
  }
  if (res < 0) res += MOD;
  cout << res << endl;
  return 0;
}

### Prompt

Generate 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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
constexpr auto MOD = 1000000009LL;
constexpr auto LINF = (1LL << 60);
constexpr auto INF = 2147483647LL;
constexpr auto PI = 3.1415926535897932384626433;
constexpr auto EPS = 1E-9;
template <typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2> p) {
  out << p.first << ' ' << p.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
auto printVector = []<typename T>(ostream& out, vector<T> v) {
  copy(v.begin(), v.end(), ostream_iterator<T>(out, " "));
};
ll pwr(ll b, ll e) {
  b %= MOD;
  if (e == 0) return 1;
  if (e == 1) return b;
  return (pwr(b * b, e / 2) * pwr(b, e % 2)) % MOD;
}
int n, a, b, k;
string str;
ll res, mve, kmove, now;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> a >> b >> k;
  string str;
  cin >> str;
  mve = pwr(a, MOD - 2) * b % MOD;
  kmove = pwr(mve, k);
  now = pwr(a, n);
  now = (kmove == 1 ? now * (n + 1) / k % MOD
                    : now * (1 - pwr(kmove, (n + 1) / k)) % MOD *
                          pwr(1 - kmove, MOD - 2) % MOD);
  for (int i = 0; i < k; i++) {
    (res += (str[i] == '+' ? now : -now)) %= MOD;
    (now *= mve) %= MOD;
  }
  if (res < 0) res += MOD;
  cout << res << endl;
  return 0;
}

```