#include <bits/stdc++.h>
using namespace std;
long long int md = 1e9 + 9;
long long int __gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}
long long int poww(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= md;
    }
    a *= a;
    a %= md;
    b >>= 1;
  }
  return res;
}
void ainp(long long int arr[], long long int n) {
  for (long long int i = 0; i < n; i++) cin >> arr[i];
}
void vinp(vector<long long int> v, long long int n) {
  for (long long int i = 0; i < n; i++) {
    long long int a;
    cin >> a;
    v.push_back(a);
  }
}
map<long long int, long long int> mapp;
set<long long int> s;
vector<long long int> v;
void fun() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
signed main() {
  fun();
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long int res = 0;
  for (long long int i = 0; i < k; i++) {
    if (s[i] == '+') {
      res = (res + poww(a, n - i) * poww(b, i) % md) % md;
    }
    if (s[i] == '-') {
      res = (res - poww(a, n - i) * poww(b, i) % md);
      res = (res + md) % md;
    }
  }
  long long int x = (n + 1) / k;
  long long int r = poww(b * poww(a, md - 2) % md, k);
  if (r == 1) {
    cout << res * ((n + 1) / k) % md;
  } else {
    res = res * poww(r - 1, md - 2) % md * (poww(r, x) - 1) % md;
    cout << res << endl;
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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 int md = 1e9 + 9;
long long int __gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return __gcd(b, a % b);
}
long long int poww(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= md;
    }
    a *= a;
    a %= md;
    b >>= 1;
  }
  return res;
}
void ainp(long long int arr[], long long int n) {
  for (long long int i = 0; i < n; i++) cin >> arr[i];
}
void vinp(vector<long long int> v, long long int n) {
  for (long long int i = 0; i < n; i++) {
    long long int a;
    cin >> a;
    v.push_back(a);
  }
}
map<long long int, long long int> mapp;
set<long long int> s;
vector<long long int> v;
void fun() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
signed main() {
  fun();
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long int res = 0;
  for (long long int i = 0; i < k; i++) {
    if (s[i] == '+') {
      res = (res + poww(a, n - i) * poww(b, i) % md) % md;
    }
    if (s[i] == '-') {
      res = (res - poww(a, n - i) * poww(b, i) % md);
      res = (res + md) % md;
    }
  }
  long long int x = (n + 1) / k;
  long long int r = poww(b * poww(a, md - 2) % md, k);
  if (r == 1) {
    cout << res * ((n + 1) / k) % md;
  } else {
    res = res * poww(r - 1, md - 2) % md * (poww(r, x) - 1) % md;
    cout << res << endl;
  }
}

```