#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
long long pw(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
      res %= m;
    }
    a = a * a;
    a %= m;
    b = b / 2;
  }
  return res;
}
long long inverse(long long first) {
  return pw(first, 1000000009 - 2, 1000000009);
}
void add(long long &a, long long b) {
  a += b;
  if (a >= 1000000009) a -= 1000000009;
  if (a < 0) a += 1000000009;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long tab[k];
  string s;
  cin >> s;
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+')
      tab[i] = 1;
    else
      tab[i] = -1;
  }
  long long ans = 0;
  for (long long i = 0; i < k; i++) {
    add(ans, (tab[i] * (pw(b, i, 1000000009) * pw(a, n - i, 1000000009)) %
              1000000009) %
                 1000000009);
  }
  n++;
  n /= k;
  long long r = ((pw(b, k, 1000000009) % 1000000009) *
                     (inverse(pw(a, k, 1000000009)) % 1000000009) +
                 1000000009) %
                1000000009;
  if (r == 1) {
    ans = ((ans % 1000000009) * (n % 1000000009) + 1000000009) % 1000000009;
    cout << ans;
    return 0;
  } else {
    long long first = pw(r, n, 1000000009);
    first--;
    if (first < 0) first += 1000000009;
    add(r, -1ll);
    long long c =
        ((first % 1000000009) * (inverse(r) % 1000000009) + 1000000009) %
        1000000009;
    ans = ((ans % 1000000009) * (c % 1000000009) + 1000000009) % 1000000009;
    cout << ans;
    return 0;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
long long pw(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a;
      res %= m;
    }
    a = a * a;
    a %= m;
    b = b / 2;
  }
  return res;
}
long long inverse(long long first) {
  return pw(first, 1000000009 - 2, 1000000009);
}
void add(long long &a, long long b) {
  a += b;
  if (a >= 1000000009) a -= 1000000009;
  if (a < 0) a += 1000000009;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long tab[k];
  string s;
  cin >> s;
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+')
      tab[i] = 1;
    else
      tab[i] = -1;
  }
  long long ans = 0;
  for (long long i = 0; i < k; i++) {
    add(ans, (tab[i] * (pw(b, i, 1000000009) * pw(a, n - i, 1000000009)) %
              1000000009) %
                 1000000009);
  }
  n++;
  n /= k;
  long long r = ((pw(b, k, 1000000009) % 1000000009) *
                     (inverse(pw(a, k, 1000000009)) % 1000000009) +
                 1000000009) %
                1000000009;
  if (r == 1) {
    ans = ((ans % 1000000009) * (n % 1000000009) + 1000000009) % 1000000009;
    cout << ans;
    return 0;
  } else {
    long long first = pw(r, n, 1000000009);
    first--;
    if (first < 0) first += 1000000009;
    add(r, -1ll);
    long long c =
        ((first % 1000000009) * (inverse(r) % 1000000009) + 1000000009) %
        1000000009;
    ans = ((ans % 1000000009) * (c % 1000000009) + 1000000009) % 1000000009;
    cout << ans;
    return 0;
  }
  return 0;
}

```