#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
const long long mod = 1e9 + 9;
long long quick_pow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1LL) res = res * a % mod;
    a = a * a % mod;
    n >>= 1LL;
  }
  return res;
}
char s[maxn];
int main() {
  int n, k;
  long long a, b;
  scanf("%d %I64d %I64d %d", &n, &a, &b, &k);
  scanf("%s", s);
  if (n < k) {
    long long ans = 0;
    for (int i = 0; i <= n; i++) {
      if (s[i] == '+') {
        ans = (ans + quick_pow(a, n - i) * quick_pow(b, i) % mod) % mod;
      } else
        ans = (ans + mod - quick_pow(a, n - i) * quick_pow(b, i) % mod) % mod;
    }
    printf("%I64d\n", ans);
  } else {
    long long af = 0, q = 0;
    for (int i = 0; i < k; i++) {
      if (s[i] == '+') {
        af = (af + quick_pow(a, n - i) * quick_pow(b, i) % mod) % mod;
      } else
        af = ((af + mod - quick_pow(a, n - i) * quick_pow(b, i) % mod)) % mod;
    }
    q = quick_pow(b, k);
    for (int i = 0; i < k; i++) q = q * quick_pow(a, mod - 2) % mod;
    long long ans = 0;
    long long cnt = n / k;
    if (q == 1LL) {
      ans = (ans + af * cnt) % mod;
    } else {
      ans = af * ((quick_pow(q, cnt) + mod - 1) % mod) % mod;
      ans = ans * quick_pow(q - 1, mod - 2) % mod;
    }
    for (int i = cnt * k; i <= n; i++) {
      if (s[i - cnt * k] == '+') {
        ans = (ans + quick_pow(a, n - i) * quick_pow(b, i) % mod) % mod;
      } else
        ans = ((ans + mod - quick_pow(a, n - i) * quick_pow(b, i) % mod)) % mod;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 int maxn = 1e5 + 50;
const long long mod = 1e9 + 9;
long long quick_pow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1LL) res = res * a % mod;
    a = a * a % mod;
    n >>= 1LL;
  }
  return res;
}
char s[maxn];
int main() {
  int n, k;
  long long a, b;
  scanf("%d %I64d %I64d %d", &n, &a, &b, &k);
  scanf("%s", s);
  if (n < k) {
    long long ans = 0;
    for (int i = 0; i <= n; i++) {
      if (s[i] == '+') {
        ans = (ans + quick_pow(a, n - i) * quick_pow(b, i) % mod) % mod;
      } else
        ans = (ans + mod - quick_pow(a, n - i) * quick_pow(b, i) % mod) % mod;
    }
    printf("%I64d\n", ans);
  } else {
    long long af = 0, q = 0;
    for (int i = 0; i < k; i++) {
      if (s[i] == '+') {
        af = (af + quick_pow(a, n - i) * quick_pow(b, i) % mod) % mod;
      } else
        af = ((af + mod - quick_pow(a, n - i) * quick_pow(b, i) % mod)) % mod;
    }
    q = quick_pow(b, k);
    for (int i = 0; i < k; i++) q = q * quick_pow(a, mod - 2) % mod;
    long long ans = 0;
    long long cnt = n / k;
    if (q == 1LL) {
      ans = (ans + af * cnt) % mod;
    } else {
      ans = af * ((quick_pow(q, cnt) + mod - 1) % mod) % mod;
      ans = ans * quick_pow(q - 1, mod - 2) % mod;
    }
    for (int i = cnt * k; i <= n; i++) {
      if (s[i - cnt * k] == '+') {
        ans = (ans + quick_pow(a, n - i) * quick_pow(b, i) % mod) % mod;
      } else
        ans = ((ans + mod - quick_pow(a, n - i) * quick_pow(b, i) % mod)) % mod;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```