#include <bits/stdc++.h>
char s[100005];
long long powa[100005], powb[100005];
long long powm(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1LL) ret = (ret * a) % 1000000009LL;
    a = (a * a) % 1000000009LL;
    b >>= 1;
  }
  return ret;
}
int main() {
  long long n, a, b, k, t, ans, p;
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", s);
  powa[k - 1] = powm(a, n - k + 1);
  powb[0] = 1;
  for (long long i = k - 2; i >= 0; i--)
    powa[i] = powa[i + 1] * a % 1000000009LL;
  for (long long i = 1; i < k; i++) powb[i] = powb[i - 1] * b % 1000000009LL;
  t = 0;
  for (long long i = 0; i < k; i++)
    if (s[i] == '+')
      t = (t + powa[i] * powb[i] % 1000000009LL) % 1000000009LL;
    else if (s[i] == '-')
      t = (t - powa[i] * powb[i] % 1000000009LL + 1000000009LL) % 1000000009LL;
  p = powm(a, 1000000009LL - 2);
  p = powm(p, k) * powm(b, k) % 1000000009LL;
  if (p == 1)
    ans = t * ((n + 1) / k) % 1000000009LL;
  else
    ans = (t * powm(p, (n + 1) / k) % 1000000009LL - t + 1000000009LL) %
          1000000009LL *
          powm((p + 1000000009LL - 1) % 1000000009LL, 1000000009LL - 2) %
          1000000009LL;
  printf("%I64d\n", ans);
  return 0;
}

### 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>
char s[100005];
long long powa[100005], powb[100005];
long long powm(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1LL) ret = (ret * a) % 1000000009LL;
    a = (a * a) % 1000000009LL;
    b >>= 1;
  }
  return ret;
}
int main() {
  long long n, a, b, k, t, ans, p;
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", s);
  powa[k - 1] = powm(a, n - k + 1);
  powb[0] = 1;
  for (long long i = k - 2; i >= 0; i--)
    powa[i] = powa[i + 1] * a % 1000000009LL;
  for (long long i = 1; i < k; i++) powb[i] = powb[i - 1] * b % 1000000009LL;
  t = 0;
  for (long long i = 0; i < k; i++)
    if (s[i] == '+')
      t = (t + powa[i] * powb[i] % 1000000009LL) % 1000000009LL;
    else if (s[i] == '-')
      t = (t - powa[i] * powb[i] % 1000000009LL + 1000000009LL) % 1000000009LL;
  p = powm(a, 1000000009LL - 2);
  p = powm(p, k) * powm(b, k) % 1000000009LL;
  if (p == 1)
    ans = t * ((n + 1) / k) % 1000000009LL;
  else
    ans = (t * powm(p, (n + 1) / k) % 1000000009LL - t + 1000000009LL) %
          1000000009LL *
          powm((p + 1000000009LL - 1) % 1000000009LL, 1000000009LL - 2) %
          1000000009LL;
  printf("%I64d\n", ans);
  return 0;
}

```