#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
const long long mod = 1000000009;
char s[N + 5];
long long power(long long x, long long a) {
  long long tmp = x;
  long long res = 1;
  while (a) {
    if (a & 1) res = res * tmp % mod;
    tmp = tmp * tmp % mod;
    a >>= 1;
  }
  return res;
}
int main() {
  long long n, a, b, k;
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", s);
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      ans = (ans + power(a, n - i) * power(b, i) % mod) % mod;
    } else
      ans = (ans - power(a, n - i) * power(b, i) % mod) % mod;
  }
  ans = (ans % mod + mod) % mod;
  long long na = power(a, mod - 2);
  long long q = power(na, k) * power(b, k) % mod;
  long long nk = (n + 1) / k;
  if (q == 1) {
    long long res = ans * nk % mod;
    printf("%I64d\n", res);
  } else {
    q = (q + mod) % mod;
    long long tmp = (power(q, nk) - 1) % mod * power(q - 1, mod - 2) % mod;
    long long res = (ans * tmp % mod + mod) % mod;
    printf("%I64d\n", res);
  }
}

### Prompt

Generate 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;
const int N = 100000;
const long long mod = 1000000009;
char s[N + 5];
long long power(long long x, long long a) {
  long long tmp = x;
  long long res = 1;
  while (a) {
    if (a & 1) res = res * tmp % mod;
    tmp = tmp * tmp % mod;
    a >>= 1;
  }
  return res;
}
int main() {
  long long n, a, b, k;
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", s);
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      ans = (ans + power(a, n - i) * power(b, i) % mod) % mod;
    } else
      ans = (ans - power(a, n - i) * power(b, i) % mod) % mod;
  }
  ans = (ans % mod + mod) % mod;
  long long na = power(a, mod - 2);
  long long q = power(na, k) * power(b, k) % mod;
  long long nk = (n + 1) / k;
  if (q == 1) {
    long long res = ans * nk % mod;
    printf("%I64d\n", res);
  } else {
    q = (q + mod) % mod;
    long long tmp = (power(q, nk) - 1) % mod * power(q - 1, mod - 2) % mod;
    long long res = (ans * tmp % mod + mod) % mod;
    printf("%I64d\n", res);
  }
}

```