#include <bits/stdc++.h>
const int MAXN = 1e5 + 10;
const int mod = 1e9 + 9;
using namespace std;
long long read() {
  long long x = 0, first = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') first = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return first * x;
}
long long ksm(long long a, long long b) {
  long long ans = 1;
  long long t = a;
  while (b) {
    if (b & 1) ans = ans * t % mod;
    b = b >> 1;
    t = t * t % mod;
  }
  return ans;
}
string second;
int main() {
  long long n, a, b, k;
  n = read();
  a = read();
  b = read();
  k = read();
  cin >> second;
  long long num = 0;
  for (int i = 0; i <= k - 1; i++) {
    if (second[i] == '+')
      num += ksm(a, n - i) * ksm(b, i) % mod;
    else
      num -= ksm(a, n - i) * ksm(b, i) % mod;
    num = (num + mod) % mod;
  }
  long long q = ksm(a, k * (mod - 2ll) % (mod - 1)) * ksm(b, k) % mod;
  long long ans = 0;
  if (q == 1)
    printf("%lld\n", (((num * ((n + 1) / k) % mod) + mod) % mod));
  else {
    printf("%lld\n", (((num * ksm(q - 1, mod - 2) % mod) *
                       (ksm(q, (n + 1) / k) - 1) % mod) +
                      mod) %
                         mod);
  }
  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>
const int MAXN = 1e5 + 10;
const int mod = 1e9 + 9;
using namespace std;
long long read() {
  long long x = 0, first = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') first = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return first * x;
}
long long ksm(long long a, long long b) {
  long long ans = 1;
  long long t = a;
  while (b) {
    if (b & 1) ans = ans * t % mod;
    b = b >> 1;
    t = t * t % mod;
  }
  return ans;
}
string second;
int main() {
  long long n, a, b, k;
  n = read();
  a = read();
  b = read();
  k = read();
  cin >> second;
  long long num = 0;
  for (int i = 0; i <= k - 1; i++) {
    if (second[i] == '+')
      num += ksm(a, n - i) * ksm(b, i) % mod;
    else
      num -= ksm(a, n - i) * ksm(b, i) % mod;
    num = (num + mod) % mod;
  }
  long long q = ksm(a, k * (mod - 2ll) % (mod - 1)) * ksm(b, k) % mod;
  long long ans = 0;
  if (q == 1)
    printf("%lld\n", (((num * ((n + 1) / k) % mod) + mod) % mod));
  else {
    printf("%lld\n", (((num * ksm(q - 1, mod - 2) % mod) *
                       (ksm(q, (n + 1) / k) - 1) % mod) +
                      mod) %
                         mod);
  }
  return 0;
}

```