#include <bits/stdc++.h>
const double Pi = acos(-1.0);
using namespace std;
const long long mod = 1000000009;
long long n, a, b, k;
string s;
long long exp(long long x, long long p) {
  if (p == 0) return 1;
  if (p & 1) return (x * exp(x, p - 1)) % mod;
  return exp((x * x) % mod, p >> 1);
}
int main(int argc, char** argv) {
  cin >> n >> a >> b >> k >> s;
  long long sum = 0;
  for (int i = 0; i < k; i++)
    sum = (sum + ((exp(a, n - i) * exp(b, i) % mod) * ((s[i] == '+') ? 1 : -1) +
                  mod) %
                     mod) %
          mod;
  long long r = (b * exp(a, mod - 2)) % mod, n1 = (n + 1) / k;
  r = exp(r, k);
  if (r == 1)
    printf("%lld", (sum * n1) % mod);
  else
    printf("%lld",
           ((((exp(r, n1) - 1) * exp(r - 1, mod - 2)) % mod) * sum) % mod);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 double Pi = acos(-1.0);
using namespace std;
const long long mod = 1000000009;
long long n, a, b, k;
string s;
long long exp(long long x, long long p) {
  if (p == 0) return 1;
  if (p & 1) return (x * exp(x, p - 1)) % mod;
  return exp((x * x) % mod, p >> 1);
}
int main(int argc, char** argv) {
  cin >> n >> a >> b >> k >> s;
  long long sum = 0;
  for (int i = 0; i < k; i++)
    sum = (sum + ((exp(a, n - i) * exp(b, i) % mod) * ((s[i] == '+') ? 1 : -1) +
                  mod) %
                     mod) %
          mod;
  long long r = (b * exp(a, mod - 2)) % mod, n1 = (n + 1) / k;
  r = exp(r, k);
  if (r == 1)
    printf("%lld", (sum * n1) % mod);
  else
    printf("%lld",
           ((((exp(r, n1) - 1) * exp(r - 1, mod - 2)) % mod) * sum) % mod);
  return 0;
}

```