#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 666;
const long long mod = 1e9 + 9;
char st[maxn];
long long quick_mod(long long a, long long n, long long mod) {
  long long temp = 1;
  while (n) {
    if (n & 1) temp = (temp * a) % mod;
    a = ((a % mod) * (a % mod)) % mod;
    n >>= 1;
  }
  return (temp % mod + mod) % mod;
}
int main() {
  int n, a, b, k, m;
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", st);
  m = n + 1;
  int x = m / k, y, chx, xx;
  long long aa, bb;
  y = m % (x * k);
  int len = k;
  long long sum = 0, ans;
  aa = quick_mod(a, k, mod);
  bb = quick_mod(b, k, mod);
  long long inv = (quick_mod(aa - bb, mod - 2, mod) + mod) % mod;
  for (int i = 0; i < len; i++) {
    chx = st[i] == '+' ? 1 : -1;
    xx = (n - i) / k;
    ans = (quick_mod(b, i, mod) * quick_mod(a, (n - i) % k, mod) + mod) % mod;
    if (aa != bb)
      ans = (((quick_mod(aa, xx + 1, mod) - quick_mod(bb, xx + 1, mod) + mod) %
              mod * inv) %
             mod) *
            ans % mod;
    else
      ans = (aa * (xx + 1)) % mod * ans % mod;
    sum = (sum + ((ans * chx) % mod) + mod) % mod;
    sum = (sum + mod) % mod;
  }
  printf("%lld\n", (sum + mod) % mod);
}

### 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;
const int maxn = 1e5 + 666;
const long long mod = 1e9 + 9;
char st[maxn];
long long quick_mod(long long a, long long n, long long mod) {
  long long temp = 1;
  while (n) {
    if (n & 1) temp = (temp * a) % mod;
    a = ((a % mod) * (a % mod)) % mod;
    n >>= 1;
  }
  return (temp % mod + mod) % mod;
}
int main() {
  int n, a, b, k, m;
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", st);
  m = n + 1;
  int x = m / k, y, chx, xx;
  long long aa, bb;
  y = m % (x * k);
  int len = k;
  long long sum = 0, ans;
  aa = quick_mod(a, k, mod);
  bb = quick_mod(b, k, mod);
  long long inv = (quick_mod(aa - bb, mod - 2, mod) + mod) % mod;
  for (int i = 0; i < len; i++) {
    chx = st[i] == '+' ? 1 : -1;
    xx = (n - i) / k;
    ans = (quick_mod(b, i, mod) * quick_mod(a, (n - i) % k, mod) + mod) % mod;
    if (aa != bb)
      ans = (((quick_mod(aa, xx + 1, mod) - quick_mod(bb, xx + 1, mod) + mod) %
              mod * inv) %
             mod) *
            ans % mod;
    else
      ans = (aa * (xx + 1)) % mod * ans % mod;
    sum = (sum + ((ans * chx) % mod) + mod) % mod;
    sum = (sum + mod) % mod;
  }
  printf("%lld\n", (sum + mod) % mod);
}

```