#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k;
long long ksm(long long x, long long t) {
  long long y = 1;
  for (; t; t >>= 1) t& 1 ? y = y * x % 1000000009 : 0, x = x * x % 1000000009;
  return y;
}
long long co, ans;
char S[100005];
signed main() {
  cin >> n >> a >> b >> k;
  co = ksm(ksm(a, 1000000009 - 2) * b % 1000000009, k);
  if (co == 1)
    co = (n + 1) / k;
  else
    co = (ksm(co, n / k + 1) - 1) * ksm(co - 1, 1000000009 - 2) % 1000000009;
  cin >> S;
  for (long long i = 0; i < k; i++)
    ans += ksm(a, n - i) * ksm(b, i) % 1000000009 * co % 1000000009 *
           (S[i] == '+' ? 1 : -1);
  cout << (ans % 1000000009 + 1000000009) % 1000000009;
}

### Prompt

Your challenge is to write 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;
long long n, a, b, k;
long long ksm(long long x, long long t) {
  long long y = 1;
  for (; t; t >>= 1) t& 1 ? y = y * x % 1000000009 : 0, x = x * x % 1000000009;
  return y;
}
long long co, ans;
char S[100005];
signed main() {
  cin >> n >> a >> b >> k;
  co = ksm(ksm(a, 1000000009 - 2) * b % 1000000009, k);
  if (co == 1)
    co = (n + 1) / k;
  else
    co = (ksm(co, n / k + 1) - 1) * ksm(co - 1, 1000000009 - 2) % 1000000009;
  cin >> S;
  for (long long i = 0; i < k; i++)
    ans += ksm(a, n - i) * ksm(b, i) % 1000000009 * co % 1000000009 *
           (S[i] == '+' ? 1 : -1);
  cout << (ans % 1000000009 + 1000000009) % 1000000009;
}

```