#include <bits/stdc++.h>
using namespace std;
string s;
long long mod = 1e9 + 9;
long long c[100005];
long long qpow(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans = (ans * x) % mod;
    }
    n /= 2;
    x = (x * x) % mod;
  }
  return ans;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> s;
  for (int i = 0; i < k; i++) {
    if (s[i] == '-') {
      c[i] = -1;
    } else {
      c[i] = 1;
    }
  }
  long long sum = 0;
  if (n < k) {
    for (int i = 0; i <= n; i++) {
      long long ss = (qpow(a, n - i) % mod) * qpow(b, i) % mod;
      if (c[i] == -1) ss = (mod - ss) % mod;
      sum = (sum + ss) % mod;
    }
    cout << (sum + mod) % mod << endl;
  } else {
    for (int i = 0; i < k; i++) {
      long long ss = qpow(a, n - i) * qpow(b, i) % mod;
      if (c[i] == -1) ss = (mod - ss) % mod;
      sum = (sum + ss) % mod;
    }
    if (qpow(a, k) == qpow(b, k)) {
      long long qq = (n + 1) / k;
      qq = qq % mod;
      sum = (sum * qq) % mod;
      cout << sum << endl;
      return 0;
    }
    long long ans1 = qpow(a, n + 1);
    long long ans2 = qpow(b, n + 1);
    long long ans3 = (ans1 - ans2 + mod) % mod * qpow(a, k) % mod;
    long long ans4 =
        (qpow(a, k) - qpow(b, k) + mod) % mod * qpow(a, n + 1) % mod;
    sum = (sum * ans3) % mod * qpow(ans4, mod - 2) % mod;
    cout << sum << endl;
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
string s;
long long mod = 1e9 + 9;
long long c[100005];
long long qpow(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans = (ans * x) % mod;
    }
    n /= 2;
    x = (x * x) % mod;
  }
  return ans;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> s;
  for (int i = 0; i < k; i++) {
    if (s[i] == '-') {
      c[i] = -1;
    } else {
      c[i] = 1;
    }
  }
  long long sum = 0;
  if (n < k) {
    for (int i = 0; i <= n; i++) {
      long long ss = (qpow(a, n - i) % mod) * qpow(b, i) % mod;
      if (c[i] == -1) ss = (mod - ss) % mod;
      sum = (sum + ss) % mod;
    }
    cout << (sum + mod) % mod << endl;
  } else {
    for (int i = 0; i < k; i++) {
      long long ss = qpow(a, n - i) * qpow(b, i) % mod;
      if (c[i] == -1) ss = (mod - ss) % mod;
      sum = (sum + ss) % mod;
    }
    if (qpow(a, k) == qpow(b, k)) {
      long long qq = (n + 1) / k;
      qq = qq % mod;
      sum = (sum * qq) % mod;
      cout << sum << endl;
      return 0;
    }
    long long ans1 = qpow(a, n + 1);
    long long ans2 = qpow(b, n + 1);
    long long ans3 = (ans1 - ans2 + mod) % mod * qpow(a, k) % mod;
    long long ans4 =
        (qpow(a, k) - qpow(b, k) + mod) % mod * qpow(a, n + 1) % mod;
    sum = (sum * ans3) % mod * qpow(ans4, mod - 2) % mod;
    cout << sum << endl;
  }
  return 0;
}

```