#include <bits/stdc++.h>
using namespace std;
void solve();
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  for (int i = 1; i <= t; ++i) solve();
  cerr << "Time taken: " << ((clock() * 1000) / CLOCKS_PER_SEC) << "ms\n";
}
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
inline long long InverseMod(long long a, long long m) {
  return pow(a, m - 2, m);
}
inline long long add(long long a, long long b, long long mod) {
  long long c = (a + b);
  if (c >= mod) {
    c -= mod;
  }
  return c;
}
inline long long sub(long long a, long long b, long long mod) {
  long long c = (a - b);
  if (c < 0) {
    c += mod;
  }
  return c;
}
long long mul(long long a, long long b, long long mod) {
  long long c = (a * b) % mod;
  return c;
}
inline long long div(long long a, long long b, long long mod) {
  if (b == 0) return a;
  long long c = mul(a, InverseMod(b, mod), mod);
  return c;
}
void solve() {
  const long long mod = 1e9 + 9;
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  std::vector<long long> v(k);
  for (long long i = 0; i < ((long long)s.size()); ++i) {
    if (s[i] == '+')
      v[i] = 1;
    else
      v[i] = -1;
  }
  long long ans = 0;
  long long len = ((n + 1) / k);
  long long num = pow(a, k * len, mod);
  num = sub(num, pow(b, k * len, mod), mod);
  num = mul(num, pow(a, n - k + 1, mod), mod);
  long long den =
      sub(pow(a, len * k, mod),
          mul(pow(a, (len - 1) * (k), mod), pow(b, k, mod), mod), mod);
  long long nd = div(num, den, mod);
  long long temp = pow(a, k - 1, mod);
  if (den == 0) {
    nd = len * pow(a, n - k + 1, mod);
  }
  for (long long i = 0; i < k; ++i) {
    if (v[i] == 1)
      ans = add(ans, temp, mod);
    else {
      ans = sub(ans, temp, mod);
    }
    temp = mul(temp, b, mod);
    temp = div(temp, a, mod);
  }
  ans = mul(ans, nd, mod);
  cout << ans << '\n';
}

### 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;
void solve();
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  for (int i = 1; i <= t; ++i) solve();
  cerr << "Time taken: " << ((clock() * 1000) / CLOCKS_PER_SEC) << "ms\n";
}
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
inline long long InverseMod(long long a, long long m) {
  return pow(a, m - 2, m);
}
inline long long add(long long a, long long b, long long mod) {
  long long c = (a + b);
  if (c >= mod) {
    c -= mod;
  }
  return c;
}
inline long long sub(long long a, long long b, long long mod) {
  long long c = (a - b);
  if (c < 0) {
    c += mod;
  }
  return c;
}
long long mul(long long a, long long b, long long mod) {
  long long c = (a * b) % mod;
  return c;
}
inline long long div(long long a, long long b, long long mod) {
  if (b == 0) return a;
  long long c = mul(a, InverseMod(b, mod), mod);
  return c;
}
void solve() {
  const long long mod = 1e9 + 9;
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  std::vector<long long> v(k);
  for (long long i = 0; i < ((long long)s.size()); ++i) {
    if (s[i] == '+')
      v[i] = 1;
    else
      v[i] = -1;
  }
  long long ans = 0;
  long long len = ((n + 1) / k);
  long long num = pow(a, k * len, mod);
  num = sub(num, pow(b, k * len, mod), mod);
  num = mul(num, pow(a, n - k + 1, mod), mod);
  long long den =
      sub(pow(a, len * k, mod),
          mul(pow(a, (len - 1) * (k), mod), pow(b, k, mod), mod), mod);
  long long nd = div(num, den, mod);
  long long temp = pow(a, k - 1, mod);
  if (den == 0) {
    nd = len * pow(a, n - k + 1, mod);
  }
  for (long long i = 0; i < k; ++i) {
    if (v[i] == 1)
      ans = add(ans, temp, mod);
    else {
      ans = sub(ans, temp, mod);
    }
    temp = mul(temp, b, mod);
    temp = div(temp, a, mod);
  }
  ans = mul(ans, nd, mod);
  cout << ans << '\n';
}

```