#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 9, maxn = 100010;
long long _pow(long long x, long long n) {
  long long ret = 1;
  for (; n; n >>= 1, x = x * x % mod)
    if (n & 1) ret = ret * x % mod;
  return ret;
}
long long n, a, b, K;
char s[maxn];
int main() {
  scanf("%I64d%I64d%I64d%I64d%s", &n, &a, &b, &K, s);
  long long inva = _pow(a, mod - 2);
  long long ret = _pow(inva, K) * _pow(b, K) % mod;
  long long sum = 0, ans = 0;
  long long afac = _pow(a, n), bfac = 1;
  for (register int i = 0; i < K; ++i) {
    if (s[i] == '+')
      sum = (sum + afac * bfac % mod) % mod;
    else
      sum = (sum + (mod - afac * bfac % mod)) % mod;
    afac = afac * inva % mod;
    bfac = bfac * b % mod;
  }
  K = (n + 1) / K;
  if (ret == 1) {
    printf("%I64d\n", sum * K % mod);
    return 0;
  }
  ans = ((mod + _pow(ret, K) - 1) % mod) *
        _pow((ret - 1 + mod) % mod, mod - 2) % mod;
  printf("%I64d\n", sum * ans % mod);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 mod = 1e9 + 9, maxn = 100010;
long long _pow(long long x, long long n) {
  long long ret = 1;
  for (; n; n >>= 1, x = x * x % mod)
    if (n & 1) ret = ret * x % mod;
  return ret;
}
long long n, a, b, K;
char s[maxn];
int main() {
  scanf("%I64d%I64d%I64d%I64d%s", &n, &a, &b, &K, s);
  long long inva = _pow(a, mod - 2);
  long long ret = _pow(inva, K) * _pow(b, K) % mod;
  long long sum = 0, ans = 0;
  long long afac = _pow(a, n), bfac = 1;
  for (register int i = 0; i < K; ++i) {
    if (s[i] == '+')
      sum = (sum + afac * bfac % mod) % mod;
    else
      sum = (sum + (mod - afac * bfac % mod)) % mod;
    afac = afac * inva % mod;
    bfac = bfac * b % mod;
  }
  K = (n + 1) / K;
  if (ret == 1) {
    printf("%I64d\n", sum * K % mod);
    return 0;
  }
  ans = ((mod + _pow(ret, K) - 1) % mod) *
        _pow((ret - 1 + mod) % mod, mod - 2) % mod;
  printf("%I64d\n", sum * ans % mod);
  return 0;
}

```