#include <bits/stdc++.h>
using namespace std;
const int Mod = 1e9 + 9;
int mul(int a, int b) { return (int)((long long)a * b % Mod); }
int pot(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a), b >>= 1;
  }
  return r;
}
int inv(int a) { return pot(a, Mod - 2); }
int n, a, b, k;
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> a >> b >> k;
  cin >> s;
  long long sol = 0;
  for (int i = (int)(0); i < (int)(k); ++i)
    sol += (s[i] == '+' ? +1 : -1) * mul(pot(a, k - 1 - i), pot(b, i));
  sol %= Mod;
  if (sol < 0) sol += Mod;
  int num = (n + 1) / k;
  int aa = pot(a, k);
  int bb = pot(b, k);
  int sol2;
  if (aa != bb)
    sol2 = mul((pot(aa, num) - pot(bb, num)), inv(aa - bb));
  else
    sol2 = mul(num, pot(aa, num - 1));
  if (sol2 < 0) sol2 += Mod;
  cout << mul((int)sol, (int)sol2) << endl;
  return 0;
}

### Prompt

Your task is to create 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;
const int Mod = 1e9 + 9;
int mul(int a, int b) { return (int)((long long)a * b % Mod); }
int pot(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a), b >>= 1;
  }
  return r;
}
int inv(int a) { return pot(a, Mod - 2); }
int n, a, b, k;
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> a >> b >> k;
  cin >> s;
  long long sol = 0;
  for (int i = (int)(0); i < (int)(k); ++i)
    sol += (s[i] == '+' ? +1 : -1) * mul(pot(a, k - 1 - i), pot(b, i));
  sol %= Mod;
  if (sol < 0) sol += Mod;
  int num = (n + 1) / k;
  int aa = pot(a, k);
  int bb = pot(b, k);
  int sol2;
  if (aa != bb)
    sol2 = mul((pot(aa, num) - pot(bb, num)), inv(aa - bb));
  else
    sol2 = mul(num, pot(aa, num - 1));
  if (sol2 < 0) sol2 += Mod;
  cout << mul((int)sol, (int)sol2) << endl;
  return 0;
}

```