#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000009;
long long expo(long long base, long long pow) {
  long long ans = 1;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans % mod;
}
long long inv(long long x) { return expo(x, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, a, b, k;
    cin >> n >> a >> b >> k;
    string s;
    cin >> s;
    long long sum = 0;
    long long fi, sc;
    for (long long int i = 0; i < k; i++) {
      fi = expo(a, n - i);
      sc = expo(b, i);
      if (s[i] == '+')
        sum += (fi * sc) % 1000000009;
      else
        sum -= (fi * sc) % 1000000009;
      sum %= 1000000009;
    }
    sum += 1000000009;
    sum %= 1000000009;
    long long ans = sum;
    long long dec = (n + 1) / k;
    long long temp = 0;
    long long d = (n + 1) / k - 1;
    long long tp = inv(a) * b;
    tp += 1000000009;
    tp %= 1000000009;
    long long ls = expo(tp, k) % 1000000009;
    if (ls == 1) {
      d %= 1000000009;
      long long nw = sum * (d + 1);
      nw %= 1000000009;
      cout << (nw + 1000000009) % 1000000009;
      return 0;
    }
    temp = ls * ((expo(ls, d) - 1 + 1000000009) % 1000000009);
    temp %= 1000000009;
    long long val = ls - 1 + 1000000009;
    val %= 1000000009;
    temp = inv(val) * temp;
    temp %= 1000000009;
    ans = sum + sum * temp;
    ans %= 1000000009;
    cout << (ans + 1000000009) % 1000000009;
    return 0;
  }
}

### Prompt

Please formulate 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 long long mod = 1000000009;
long long expo(long long base, long long pow) {
  long long ans = 1;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans % mod;
}
long long inv(long long x) { return expo(x, mod - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, a, b, k;
    cin >> n >> a >> b >> k;
    string s;
    cin >> s;
    long long sum = 0;
    long long fi, sc;
    for (long long int i = 0; i < k; i++) {
      fi = expo(a, n - i);
      sc = expo(b, i);
      if (s[i] == '+')
        sum += (fi * sc) % 1000000009;
      else
        sum -= (fi * sc) % 1000000009;
      sum %= 1000000009;
    }
    sum += 1000000009;
    sum %= 1000000009;
    long long ans = sum;
    long long dec = (n + 1) / k;
    long long temp = 0;
    long long d = (n + 1) / k - 1;
    long long tp = inv(a) * b;
    tp += 1000000009;
    tp %= 1000000009;
    long long ls = expo(tp, k) % 1000000009;
    if (ls == 1) {
      d %= 1000000009;
      long long nw = sum * (d + 1);
      nw %= 1000000009;
      cout << (nw + 1000000009) % 1000000009;
      return 0;
    }
    temp = ls * ((expo(ls, d) - 1 + 1000000009) % 1000000009);
    temp %= 1000000009;
    long long val = ls - 1 + 1000000009;
    val %= 1000000009;
    temp = inv(val) * temp;
    temp %= 1000000009;
    ans = sum + sum * temp;
    ans %= 1000000009;
    cout << (ans + 1000000009) % 1000000009;
    return 0;
  }
}

```