#include <bits/stdc++.h>
using namespace std;
inline long long modpow(long long base, long long exp) {
  long long res = 1;
  while (exp > 0) {
    if (exp & 1) res = (res % 1000000009 * base % 1000000009) % 1000000009;
    base = (base % 1000000009 * base % 1000000009) % 1000000009;
    exp /= 2;
  }
  return res;
}
int main() {
  long long n, a, b, k;
  scanf("%lld", &n);
  scanf("%lld", &a);
  scanf("%lld", &b);
  scanf("%lld", &k);
  string str;
  cin >> str;
  vector<long long> v;
  for (long long i = 0; i < k; i++) {
    if (str[i] == '+') {
      long long foo = modpow(a, n - i);
      long long bar = modpow(b, i);
      long long ans = (foo * bar) % 1000000009;
      v.push_back(ans);
    } else {
      long long foo = modpow(a, n - i);
      long long bar = modpow(b, i);
      long long ans = (foo * bar) % 1000000009;
      v.push_back(-ans);
    }
  }
  long long foc;
  long long den =
      (modpow(b, k) * modpow(modpow(a, k), 1000000009 - 2)) % 1000000009;
  if (den < 0) den += 1000000009;
  if (den == 1)
    foc = (n + 1) / k;
  else {
    long long num =
        (modpow(b, n + 1) * modpow(modpow(a, n + 1), 1000000009 - 2)) %
        1000000009;
    num = 1 - num;
    den = 1 - den;
    foc = (num * modpow(den, 1000000009 - 2)) % 1000000009;
  }
  for (long long i = 0; i < k; i++) {
    v[i] = (v[i] * foc) % 1000000009;
  }
  long long ans = 0;
  for (long long i = 0; i < k; i++) {
    ans += v[i];
    if (ans >= 1000000009)
      ans -= 1000000009;
    else if (ans < 0)
      ans += 1000000009;
  }
  printf("%lld\n", ans);
}

### Prompt

Please formulate 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;
inline long long modpow(long long base, long long exp) {
  long long res = 1;
  while (exp > 0) {
    if (exp & 1) res = (res % 1000000009 * base % 1000000009) % 1000000009;
    base = (base % 1000000009 * base % 1000000009) % 1000000009;
    exp /= 2;
  }
  return res;
}
int main() {
  long long n, a, b, k;
  scanf("%lld", &n);
  scanf("%lld", &a);
  scanf("%lld", &b);
  scanf("%lld", &k);
  string str;
  cin >> str;
  vector<long long> v;
  for (long long i = 0; i < k; i++) {
    if (str[i] == '+') {
      long long foo = modpow(a, n - i);
      long long bar = modpow(b, i);
      long long ans = (foo * bar) % 1000000009;
      v.push_back(ans);
    } else {
      long long foo = modpow(a, n - i);
      long long bar = modpow(b, i);
      long long ans = (foo * bar) % 1000000009;
      v.push_back(-ans);
    }
  }
  long long foc;
  long long den =
      (modpow(b, k) * modpow(modpow(a, k), 1000000009 - 2)) % 1000000009;
  if (den < 0) den += 1000000009;
  if (den == 1)
    foc = (n + 1) / k;
  else {
    long long num =
        (modpow(b, n + 1) * modpow(modpow(a, n + 1), 1000000009 - 2)) %
        1000000009;
    num = 1 - num;
    den = 1 - den;
    foc = (num * modpow(den, 1000000009 - 2)) % 1000000009;
  }
  for (long long i = 0; i < k; i++) {
    v[i] = (v[i] * foc) % 1000000009;
  }
  long long ans = 0;
  for (long long i = 0; i < k; i++) {
    ans += v[i];
    if (ans >= 1000000009)
      ans -= 1000000009;
    else if (ans < 0)
      ans += 1000000009;
  }
  printf("%lld\n", ans);
}

```