#include <bits/stdc++.h>
using namespace std;
const long long K = 1e5 + 10, P = 1e9 + 9;
long long n, a, b, k, s[K];
char S[K];
long long powa[K], powb[K], ans;
long long pow(long long a, long long x) {
  long long ret = 1;
  for (; x; a = a * a % P, x >>= 1)
    if (x & 1) ret = ret * a % P;
  return ret;
}
long long sum(long long a, long long b, long long k) {
  if (k == 1) return 1;
  long long ret =
      sum(a, b, k / 2) *
      (pow(a, k - (k / 2)) + pow(b, k / 2) * pow(a, k - 2 * (k / 2)) % P) % P;
  if (k % 2) {
    ret = (ret + pow(b, k - 1)) % P;
  }
  return ret;
}
signed main() {
  cin >> n >> a >> b >> k;
  scanf("%s", S);
  powa[0] = powb[0] = 1;
  for (long long i = (1); i < (k + 1); ++i) {
    powa[i] = powa[i - 1] * a % P;
    powb[i] = powb[i - 1] * b % P;
  }
  for (long long i = (0); i < (k); ++i) {
    s[i] = S[i] == '+' ? 1 : -1;
    ans = (P + (ans + s[i] * powa[k - 1 - i] % P * powb[i] % P) % P) % P;
  }
  ans = ans * sum(powa[k], powb[k], (n + 1) / k) % P;
  printf("%lld\n", ans);
}

### Prompt

Your task is to create 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 K = 1e5 + 10, P = 1e9 + 9;
long long n, a, b, k, s[K];
char S[K];
long long powa[K], powb[K], ans;
long long pow(long long a, long long x) {
  long long ret = 1;
  for (; x; a = a * a % P, x >>= 1)
    if (x & 1) ret = ret * a % P;
  return ret;
}
long long sum(long long a, long long b, long long k) {
  if (k == 1) return 1;
  long long ret =
      sum(a, b, k / 2) *
      (pow(a, k - (k / 2)) + pow(b, k / 2) * pow(a, k - 2 * (k / 2)) % P) % P;
  if (k % 2) {
    ret = (ret + pow(b, k - 1)) % P;
  }
  return ret;
}
signed main() {
  cin >> n >> a >> b >> k;
  scanf("%s", S);
  powa[0] = powb[0] = 1;
  for (long long i = (1); i < (k + 1); ++i) {
    powa[i] = powa[i - 1] * a % P;
    powb[i] = powb[i - 1] * b % P;
  }
  for (long long i = (0); i < (k); ++i) {
    s[i] = S[i] == '+' ? 1 : -1;
    ans = (P + (ans + s[i] * powa[k - 1 - i] % P * powb[i] % P) % P) % P;
  }
  ans = ans * sum(powa[k], powb[k], (n + 1) / k) % P;
  printf("%lld\n", ans);
}

```