#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-8;
const double pi = acos(-1.0);
const int mod = 1000000009;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 20;
long long n, a, b, k;
string s;
long long head, comm, cnt;
long long pow_mod(long long a, long long n, long long m) {
  if (n == 0) return 1;
  long long x = pow_mod(a, n / 2, m);
  long long ans = x * x % m;
  if (n & 1) ans = ans * a % mod;
  return ans;
}
void exgcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inv(long long a, long long p) {
  long long d, x, y;
  exgcd(a, p, d, x, y);
  return d == 1 ? (x + p) % p : -1;
}
int main() {
  cin >> n >> a >> b >> k >> s;
  for (int i = 0; i < k; i++) {
    long long tmp = pow_mod(a, n - i, mod) * pow_mod(b, i, mod);
    tmp %= mod;
    if (s[i] == '+') {
      head = (head + tmp + mod) % mod;
    } else {
      head = (head - tmp + mod) % mod;
    }
  }
  comm = pow_mod((inv(a, mod) * b) % mod, k, mod);
  long long cnt = (n + 1) / k;
  if (comm == 1) {
    cout << (head * cnt + mod) % mod << endl;
  } else {
    cout << (((head * ((pow_mod(comm, cnt, mod) - 1) % mod) % mod) *
              inv(comm - 1, mod)) %
                 mod +
             mod) %
                mod;
  }
  return 0;
}

### 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;
const long double eps = 1e-8;
const double pi = acos(-1.0);
const int mod = 1000000009;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 20;
long long n, a, b, k;
string s;
long long head, comm, cnt;
long long pow_mod(long long a, long long n, long long m) {
  if (n == 0) return 1;
  long long x = pow_mod(a, n / 2, m);
  long long ans = x * x % m;
  if (n & 1) ans = ans * a % mod;
  return ans;
}
void exgcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inv(long long a, long long p) {
  long long d, x, y;
  exgcd(a, p, d, x, y);
  return d == 1 ? (x + p) % p : -1;
}
int main() {
  cin >> n >> a >> b >> k >> s;
  for (int i = 0; i < k; i++) {
    long long tmp = pow_mod(a, n - i, mod) * pow_mod(b, i, mod);
    tmp %= mod;
    if (s[i] == '+') {
      head = (head + tmp + mod) % mod;
    } else {
      head = (head - tmp + mod) % mod;
    }
  }
  comm = pow_mod((inv(a, mod) * b) % mod, k, mod);
  long long cnt = (n + 1) / k;
  if (comm == 1) {
    cout << (head * cnt + mod) % mod << endl;
  } else {
    cout << (((head * ((pow_mod(comm, cnt, mod) - 1) % mod) % mod) *
              inv(comm - 1, mod)) %
                 mod +
             mod) %
                mod;
  }
  return 0;
}

```