#include <bits/stdc++.h>
using namespace std;
const long long md = 1000000009;
long long n, a, b, k;
string str;
long long arr[100010];
long long bigmod(long long p, long long q) {
  if (q == 0)
    return 1LL;
  else if (q == 1)
    return p % md;
  long long ret = bigmod(p, q / 2);
  ret = (ret * ret) % md;
  if (q & 1) ret = (ret * p) % md;
  return ret;
}
long long func(long long p, long long xx) {
  if (p == 0) return 1LL;
  if (p == 1) {
    return (bigmod(a, ((p + 1) / 2) * xx) + bigmod(b, ((p + 1) / 2) * xx)) % md;
  }
  if (p & 1) {
    return ((bigmod(a, ((p + 1) / 2) * xx) +
             bigmod(b, ((p + 1) / 2) * xx) % md) *
            func(p / 2, xx)) %
           md;
  } else {
    return ((bigmod(a, ((p + 1) / 2) * xx) +
             bigmod(b, ((p + 1) / 2) * xx) % md) *
                func(p / 2, xx) -
            (bigmod(a, ((p + 1) / 2) * xx) * bigmod(b, ((p + 1) / 2) * xx) %
             md) %
                md +
            md) %
           md;
  }
}
int main() {
  memset(arr, 0, sizeof(arr));
  cin >> n >> a >> b >> k;
  cin >> str;
  for (int i = 0; i < k; i++)
    if (str[i] == '-') arr[i] = 1;
  long long ans = func(n, 1);
  for (long long i = 0; i < k; i++) {
    if (arr[i]) {
      ans -= (2 * ((((bigmod(a, k - i - 1) * bigmod(b, i)) % md) *
                    (func((n + 1) / k - 1, k) % md)) %
                   md)) %
             md;
      ans = (ans + md) % md;
    }
  }
  cout << ans << endl;
}

### 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 long md = 1000000009;
long long n, a, b, k;
string str;
long long arr[100010];
long long bigmod(long long p, long long q) {
  if (q == 0)
    return 1LL;
  else if (q == 1)
    return p % md;
  long long ret = bigmod(p, q / 2);
  ret = (ret * ret) % md;
  if (q & 1) ret = (ret * p) % md;
  return ret;
}
long long func(long long p, long long xx) {
  if (p == 0) return 1LL;
  if (p == 1) {
    return (bigmod(a, ((p + 1) / 2) * xx) + bigmod(b, ((p + 1) / 2) * xx)) % md;
  }
  if (p & 1) {
    return ((bigmod(a, ((p + 1) / 2) * xx) +
             bigmod(b, ((p + 1) / 2) * xx) % md) *
            func(p / 2, xx)) %
           md;
  } else {
    return ((bigmod(a, ((p + 1) / 2) * xx) +
             bigmod(b, ((p + 1) / 2) * xx) % md) *
                func(p / 2, xx) -
            (bigmod(a, ((p + 1) / 2) * xx) * bigmod(b, ((p + 1) / 2) * xx) %
             md) %
                md +
            md) %
           md;
  }
}
int main() {
  memset(arr, 0, sizeof(arr));
  cin >> n >> a >> b >> k;
  cin >> str;
  for (int i = 0; i < k; i++)
    if (str[i] == '-') arr[i] = 1;
  long long ans = func(n, 1);
  for (long long i = 0; i < k; i++) {
    if (arr[i]) {
      ans -= (2 * ((((bigmod(a, k - i - 1) * bigmod(b, i)) % md) *
                    (func((n + 1) / k - 1, k) % md)) %
                   md)) %
             md;
      ans = (ans + md) % md;
    }
  }
  cout << ans << endl;
}

```