#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int fexp(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000009;
    b /= 2;
    a = a * a % 1000000009;
  }
  return ans;
}
long long int inverse(long long int a, long long int p) {
  return fexp(a, p - 2);
}
void ingraph(vector<vector<long long int> >& graph, long long int m) {
  long long int x, y;
  for (long long int i = 0; i < m; i++) {
    cin >> x >> y;
    x--, y--;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
}
long long int modify(long long int n) {
  long long int res = n;
  res %= 1000000009;
  res += 1000000009;
  res %= 1000000009;
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    long long int n, a, b, k;
    cin >> n >> a >> b >> k;
    vector<long long int> vec(k);
    char tmp;
    for (long long int i = 0; i < k; i++) {
      cin >> tmp;
      if (tmp == '+')
        vec[i] = 1;
      else
        vec[i] = -1;
    }
    long long int S = 0;
    for (long long int i = 0; i < k; i++) {
      S += (((vec[i] * fexp(a, n - i)) % 1000000009) * fexp(b, i)) % 1000000009;
      S = modify(S);
    }
    S = modify(S);
    long long int x = fexp(b, k) * inverse(fexp(a, k), 1000000009);
    x %= 1000000009;
    long long int c = fexp(x, (n + 1) / k) - 1;
    c = modify(c);
    long long int d = x - 1;
    d = modify(d);
    if (d == 0) {
      S *= (n + 1) / k;
      S = modify(S);
    } else {
      d = inverse(d, 1000000009);
      S *= c;
      S %= 1000000009;
      S *= d;
      S %= 1000000009;
    }
    S = modify(S);
    cout << S << "\n";
  }
}

### 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;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int fexp(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000009;
    b /= 2;
    a = a * a % 1000000009;
  }
  return ans;
}
long long int inverse(long long int a, long long int p) {
  return fexp(a, p - 2);
}
void ingraph(vector<vector<long long int> >& graph, long long int m) {
  long long int x, y;
  for (long long int i = 0; i < m; i++) {
    cin >> x >> y;
    x--, y--;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
}
long long int modify(long long int n) {
  long long int res = n;
  res %= 1000000009;
  res += 1000000009;
  res %= 1000000009;
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    long long int n, a, b, k;
    cin >> n >> a >> b >> k;
    vector<long long int> vec(k);
    char tmp;
    for (long long int i = 0; i < k; i++) {
      cin >> tmp;
      if (tmp == '+')
        vec[i] = 1;
      else
        vec[i] = -1;
    }
    long long int S = 0;
    for (long long int i = 0; i < k; i++) {
      S += (((vec[i] * fexp(a, n - i)) % 1000000009) * fexp(b, i)) % 1000000009;
      S = modify(S);
    }
    S = modify(S);
    long long int x = fexp(b, k) * inverse(fexp(a, k), 1000000009);
    x %= 1000000009;
    long long int c = fexp(x, (n + 1) / k) - 1;
    c = modify(c);
    long long int d = x - 1;
    d = modify(d);
    if (d == 0) {
      S *= (n + 1) / k;
      S = modify(S);
    } else {
      d = inverse(d, 1000000009);
      S *= c;
      S %= 1000000009;
      S *= d;
      S %= 1000000009;
    }
    S = modify(S);
    cout << S << "\n";
  }
}

```