#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 9;
int pot(int a, int b, int m = MOD) {
  if (b == 0) return 1;
  int q = pot(a, b / 2, m);
  int q2 = (1LL * q * q) % m;
  if (b % 2)
    return (1LL * q2 * a) % m;
  else
    return q2;
}
int inv(int first) { return pot(first, MOD - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  vector<int> A(k), B(k);
  A[0] = pot(a, n);
  B[0] = 1;
  for (int i = 1; i < k; i++) {
    A[i] = (1LL * A[i - 1] * inv(a)) % MOD;
    B[i] = (1LL * B[i - 1] * b) % MOD;
  }
  int tmp = 0;
  for (int i = 0; i < k; i++) {
    tmp += ((s[i] == '+' ? 1LL : -1LL) * A[i] * B[i]) % MOD;
    tmp %= MOD;
  }
  int q = (1LL * pot(inv(a), k) * pot(b, k)) % MOD;
  int gr = (n + 1) / k;
  if (q == 1) {
    int ret = 1LL * tmp * gr % MOD;
    ret += MOD;
    ret %= MOD;
    cout << ret << "\n";
  } else {
    int ret = ((1LL * tmp * (1 - pot(q, gr))) % MOD * inv(1 - q)) % MOD;
    ret += MOD;
    ret %= MOD;
    cout << ret << "\n";
  }
  return 0;
}

### 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;
const int MOD = 1e9 + 9;
int pot(int a, int b, int m = MOD) {
  if (b == 0) return 1;
  int q = pot(a, b / 2, m);
  int q2 = (1LL * q * q) % m;
  if (b % 2)
    return (1LL * q2 * a) % m;
  else
    return q2;
}
int inv(int first) { return pot(first, MOD - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  vector<int> A(k), B(k);
  A[0] = pot(a, n);
  B[0] = 1;
  for (int i = 1; i < k; i++) {
    A[i] = (1LL * A[i - 1] * inv(a)) % MOD;
    B[i] = (1LL * B[i - 1] * b) % MOD;
  }
  int tmp = 0;
  for (int i = 0; i < k; i++) {
    tmp += ((s[i] == '+' ? 1LL : -1LL) * A[i] * B[i]) % MOD;
    tmp %= MOD;
  }
  int q = (1LL * pot(inv(a), k) * pot(b, k)) % MOD;
  int gr = (n + 1) / k;
  if (q == 1) {
    int ret = 1LL * tmp * gr % MOD;
    ret += MOD;
    ret %= MOD;
    cout << ret << "\n";
  } else {
    int ret = ((1LL * tmp * (1 - pot(q, gr))) % MOD * inv(1 - q)) % MOD;
    ret += MOD;
    ret %= MOD;
    cout << ret << "\n";
  }
  return 0;
}

```