#include <bits/stdc++.h>
using namespace std;
long long pow2(long long x, long long y) {
  if (y == 0) return 1;
  long long tmp = pow2(x, y / 2);
  tmp = (tmp * tmp) % 1000000009;
  if (y & 1) tmp = (tmp * x) % 1000000009;
  return tmp;
}
long long inv(long long x) { return pow2(x, 1000000009 - 2); }
long long n, a, b, k;
string s;
void code() {
  cin >> n >> a >> b >> k >> s;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long tmp = pow2(a, n - i) * pow2(b, i);
    tmp %= 1000000009;
    if (s[i] == '+')
      ans += tmp;
    else
      ans -= tmp;
    ans %= 1000000009;
  }
  long long x = (n + 1) / k;
  long long tmp = x % 1000000009;
  long long bak = (b * inv(a)) % 1000000009;
  bak = pow2(bak, k);
  if (bak != 1) {
    tmp = (pow2(bak, x) - 1 + 1000000009) % 1000000009;
    tmp = (tmp * inv((bak - 1 + 1000000009) % 1000000009)) % 1000000009;
  }
  ans = (ans * tmp) % 1000000009;
  if (ans < 0) ans = (ans + 1000000009) % 1000000009;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    code();
  }
}

### 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;
long long pow2(long long x, long long y) {
  if (y == 0) return 1;
  long long tmp = pow2(x, y / 2);
  tmp = (tmp * tmp) % 1000000009;
  if (y & 1) tmp = (tmp * x) % 1000000009;
  return tmp;
}
long long inv(long long x) { return pow2(x, 1000000009 - 2); }
long long n, a, b, k;
string s;
void code() {
  cin >> n >> a >> b >> k >> s;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long tmp = pow2(a, n - i) * pow2(b, i);
    tmp %= 1000000009;
    if (s[i] == '+')
      ans += tmp;
    else
      ans -= tmp;
    ans %= 1000000009;
  }
  long long x = (n + 1) / k;
  long long tmp = x % 1000000009;
  long long bak = (b * inv(a)) % 1000000009;
  bak = pow2(bak, k);
  if (bak != 1) {
    tmp = (pow2(bak, x) - 1 + 1000000009) % 1000000009;
    tmp = (tmp * inv((bak - 1 + 1000000009) % 1000000009)) % 1000000009;
  }
  ans = (ans * tmp) % 1000000009;
  if (ans < 0) ans = (ans + 1000000009) % 1000000009;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    code();
  }
}

```