#include <bits/stdc++.h>
using namespace std;
int add(int a, int b) { return (a + b) % 1000000009; }
int sub(int a, int b) { return (a - b + 1000000009) % 1000000009; }
int mul(int a, int b) {
  return (int)(((long long)(a % 1000000009) * (b % 1000000009)) % 1000000009);
}
int divine(int a, int p = 1000000009, int b = -1) {
  if (b < 0) b = p - 2;
  if (b == 0)
    return 1;
  else if (b % 2 == 0) {
    int d = divine(a, p, b / 2);
    return mul(d, d);
  } else {
    return mul(a, divine(a, p, b - 1));
  }
}
int modpow(int a, int b) {
  if (b == 0) return 1;
  int temp = modpow(a, b / 2);
  if (b % 2 == 1)
    return mul(mul(temp, temp), a);
  else {
    return mul(temp, temp);
  }
}
int main() {
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  n++;
  string s;
  cin >> s;
  if (n <= k) {
    int ans = 0;
    for (int(i) = 0; (i) < (n); (i)++) {
      if (s[i] == '+')
        ans = add(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
      else
        ans = sub(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
    }
    cout << ans << endl;
    return 0;
  }
  int ans = 0;
  for (int(i) = 0; (i) < (k); (i)++) {
    if (s[i] == '+')
      ans = add(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
    else
      ans = sub(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
  }
  int t = n / k;
  int kouhi = mul(modpow(b, k), divine(modpow(a, k)));
  if (kouhi == 1)
    ans = mul(ans, t);
  else
    ans = mul(ans, mul(sub(1, modpow(kouhi, t)), divine(sub(1, kouhi))));
  for (int(i) = k * t; (i) < (n); (i)++) {
    if (s[i] == '+')
      ans = add(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
    else
      ans = sub(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
int add(int a, int b) { return (a + b) % 1000000009; }
int sub(int a, int b) { return (a - b + 1000000009) % 1000000009; }
int mul(int a, int b) {
  return (int)(((long long)(a % 1000000009) * (b % 1000000009)) % 1000000009);
}
int divine(int a, int p = 1000000009, int b = -1) {
  if (b < 0) b = p - 2;
  if (b == 0)
    return 1;
  else if (b % 2 == 0) {
    int d = divine(a, p, b / 2);
    return mul(d, d);
  } else {
    return mul(a, divine(a, p, b - 1));
  }
}
int modpow(int a, int b) {
  if (b == 0) return 1;
  int temp = modpow(a, b / 2);
  if (b % 2 == 1)
    return mul(mul(temp, temp), a);
  else {
    return mul(temp, temp);
  }
}
int main() {
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  n++;
  string s;
  cin >> s;
  if (n <= k) {
    int ans = 0;
    for (int(i) = 0; (i) < (n); (i)++) {
      if (s[i] == '+')
        ans = add(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
      else
        ans = sub(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
    }
    cout << ans << endl;
    return 0;
  }
  int ans = 0;
  for (int(i) = 0; (i) < (k); (i)++) {
    if (s[i] == '+')
      ans = add(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
    else
      ans = sub(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
  }
  int t = n / k;
  int kouhi = mul(modpow(b, k), divine(modpow(a, k)));
  if (kouhi == 1)
    ans = mul(ans, t);
  else
    ans = mul(ans, mul(sub(1, modpow(kouhi, t)), divine(sub(1, kouhi))));
  for (int(i) = k * t; (i) < (n); (i)++) {
    if (s[i] == '+')
      ans = add(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
    else
      ans = sub(ans, mul(modpow(a, n - i - 1), modpow(b, i)));
  }
  cout << ans << endl;
  return 0;
}

```