#include <bits/stdc++.h>
using namespace std;
int x[100010];
long long f(long long y, int n) {
  if (n == 0) return 1ll;
  long long ans = f(y, n / 2);
  long long s = 1;
  if (n % 2 == 1) {
    s = (((ans * ans) % 1000000009ll) * (long long)y) % 1000000009ll;
  } else {
    s = (ans * ans) % 1000000009ll;
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, a, b, k, s = 0, ans = 0, kol = 0;
  char c;
  cin >> n >> a >> b >> k;
  for (int i = 0; i < k; i++) {
    cin >> c;
    x[i] = ((c == '+') ? 1 : -1);
    s += (((f(a, k - i - 1) * f(b, i)) % 1000000009ll) * (long long)x[i]) %
         1000000009ll;
    s += 1000000009ll;
    s %= 1000000009ll;
  }
  int q = sqrt(n), p = k;
  while (p < q) {
    p = p + k;
  }
  s = 0;
  for (int i = 0; i < p; i++) {
    s += (((f(a, p - i - 1) * f(b, i)) % 1000000009ll) * (long long)x[i % k]) %
         1000000009ll;
    s += 1000000009ll;
    s %= 1000000009ll;
  }
  int t = (n + 1) / p;
  t = t * p;
  for (int i = 1; i <= (n + 1) / p; i++) {
    ans += (((f(a, n - i * p + 1) * f(b, (i - 1) * p)) % 1000000009ll) * s) %
           1000000009ll;
    ans += 1000000009ll;
    ans %= 1000000009ll;
  }
  for (int i = t; i <= n; i++) {
    ans += (((f(a, n - i) * f(b, i)) % 1000000009ll) * (long long)x[i % k]) %
           1000000009ll;
    ans += 1000000009ll;
    ans %= 1000000009ll;
  }
  cout << ans;
}

### Prompt

Please formulate 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;
int x[100010];
long long f(long long y, int n) {
  if (n == 0) return 1ll;
  long long ans = f(y, n / 2);
  long long s = 1;
  if (n % 2 == 1) {
    s = (((ans * ans) % 1000000009ll) * (long long)y) % 1000000009ll;
  } else {
    s = (ans * ans) % 1000000009ll;
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, a, b, k, s = 0, ans = 0, kol = 0;
  char c;
  cin >> n >> a >> b >> k;
  for (int i = 0; i < k; i++) {
    cin >> c;
    x[i] = ((c == '+') ? 1 : -1);
    s += (((f(a, k - i - 1) * f(b, i)) % 1000000009ll) * (long long)x[i]) %
         1000000009ll;
    s += 1000000009ll;
    s %= 1000000009ll;
  }
  int q = sqrt(n), p = k;
  while (p < q) {
    p = p + k;
  }
  s = 0;
  for (int i = 0; i < p; i++) {
    s += (((f(a, p - i - 1) * f(b, i)) % 1000000009ll) * (long long)x[i % k]) %
         1000000009ll;
    s += 1000000009ll;
    s %= 1000000009ll;
  }
  int t = (n + 1) / p;
  t = t * p;
  for (int i = 1; i <= (n + 1) / p; i++) {
    ans += (((f(a, n - i * p + 1) * f(b, (i - 1) * p)) % 1000000009ll) * s) %
           1000000009ll;
    ans += 1000000009ll;
    ans %= 1000000009ll;
  }
  for (int i = t; i <= n; i++) {
    ans += (((f(a, n - i) * f(b, i)) % 1000000009ll) * (long long)x[i % k]) %
           1000000009ll;
    ans += 1000000009ll;
    ans %= 1000000009ll;
  }
  cout << ans;
}

```