#include <bits/stdc++.h>
using namespace std;
long long sum;
long long a, b, n, k;
long long powr(long long m, long long n) {
  if (m == 1) return 1;
  if (n == 0) return 1;
  if (n == 1) return m % 1000000009;
  if (n % 2) {
    long long te = powr(m, n / 2);
    return (((te * te) % 1000000009) * (m % 1000000009)) % 1000000009;
  } else {
    long long te = powr(m, n / 2);
    return (te * te) % 1000000009;
  }
}
long long solve(long long go) {
  if (go == 1) return sum;
  if (go % 2) {
    long long tmp = solve(go / 2);
    long long prt = powr(a, (go - 1) * k);
    long long t = sum * powr(b, (go - 1) * k) % 1000000009;
    t = t * (powr(prt, 1000000009 - 2)) % 1000000009;
    prt = powr(a, (go / 2) * k);
    long long t2 = tmp * powr(b, (go / 2) * k) % 1000000009;
    t2 = t2 * (powr(prt, 1000000009 - 2)) % 1000000009;
    return (tmp + t + t2) % 1000000009;
  } else {
    long long tmp = solve(go / 2);
    long long prt = powr(a, (go / 2) * k);
    long long t2 = tmp * powr(b, (go / 2) * k) % 1000000009;
    t2 = t2 * (powr(prt, 1000000009 - 2)) % 1000000009;
    return (tmp + t2) % 1000000009;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    cin >> n >> a >> b >> k;
    string s;
    cin >> s;
    int ap = n, bp = 0;
    int f = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '+')
        f = 1;
      else
        f = -1;
      long long tmp;
      tmp = (powr(a, ap) * powr(b, bp)) % 1000000009;
      tmp *= f;
      sum += tmp;
      if (sum < 0) sum += 1000000009;
      sum %= 1000000009;
      ap--;
      bp++;
    }
    long long go = (n + 1) / k;
    cout << solve(go);
  }
}

### 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;
long long sum;
long long a, b, n, k;
long long powr(long long m, long long n) {
  if (m == 1) return 1;
  if (n == 0) return 1;
  if (n == 1) return m % 1000000009;
  if (n % 2) {
    long long te = powr(m, n / 2);
    return (((te * te) % 1000000009) * (m % 1000000009)) % 1000000009;
  } else {
    long long te = powr(m, n / 2);
    return (te * te) % 1000000009;
  }
}
long long solve(long long go) {
  if (go == 1) return sum;
  if (go % 2) {
    long long tmp = solve(go / 2);
    long long prt = powr(a, (go - 1) * k);
    long long t = sum * powr(b, (go - 1) * k) % 1000000009;
    t = t * (powr(prt, 1000000009 - 2)) % 1000000009;
    prt = powr(a, (go / 2) * k);
    long long t2 = tmp * powr(b, (go / 2) * k) % 1000000009;
    t2 = t2 * (powr(prt, 1000000009 - 2)) % 1000000009;
    return (tmp + t + t2) % 1000000009;
  } else {
    long long tmp = solve(go / 2);
    long long prt = powr(a, (go / 2) * k);
    long long t2 = tmp * powr(b, (go / 2) * k) % 1000000009;
    t2 = t2 * (powr(prt, 1000000009 - 2)) % 1000000009;
    return (tmp + t2) % 1000000009;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    cin >> n >> a >> b >> k;
    string s;
    cin >> s;
    int ap = n, bp = 0;
    int f = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == '+')
        f = 1;
      else
        f = -1;
      long long tmp;
      tmp = (powr(a, ap) * powr(b, bp)) % 1000000009;
      tmp *= f;
      sum += tmp;
      if (sum < 0) sum += 1000000009;
      sum %= 1000000009;
      ap--;
      bp++;
    }
    long long go = (n + 1) / k;
    cout << solve(go);
  }
}

```