#include <bits/stdc++.h>
using namespace std;
const long long N = 100005;
const long long MOD = 1e9 + 9;
long long n, a, b, k;
char ss[N];
long long pow(long long x, long long y) {
  if (y == 1) return x;
  long long lalal = pow(x, y >> 1);
  lalal = lalal * lalal % MOD;
  if (y & 1) lalal = lalal * x % MOD;
  return lalal;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  long long ny = pow(a, MOD - 2), now = pow(a, n), lalal = 0;
  scanf("%s", ss);
  for (long long u = 0; u < k; u++) {
    if (ss[u] == '+')
      lalal = lalal + now;
    else
      lalal = lalal - now;
    lalal = (lalal % MOD + MOD) % MOD;
    now = now * b % MOD * ny % MOD;
  }
  long long p = pow(b * ny % MOD, k), ans = 0;
  if (p != 1)
    ans = (pow(p, (n + 1) / k) - 1) * pow(p - 1, MOD - 2) % MOD * lalal % MOD;
  else
    ans = lalal * ((n + 1) / k) % MOD;
  printf("%I64d", ans);
  return 0;
}

### Prompt

Please formulate 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 long long N = 100005;
const long long MOD = 1e9 + 9;
long long n, a, b, k;
char ss[N];
long long pow(long long x, long long y) {
  if (y == 1) return x;
  long long lalal = pow(x, y >> 1);
  lalal = lalal * lalal % MOD;
  if (y & 1) lalal = lalal * x % MOD;
  return lalal;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  long long ny = pow(a, MOD - 2), now = pow(a, n), lalal = 0;
  scanf("%s", ss);
  for (long long u = 0; u < k; u++) {
    if (ss[u] == '+')
      lalal = lalal + now;
    else
      lalal = lalal - now;
    lalal = (lalal % MOD + MOD) % MOD;
    now = now * b % MOD * ny % MOD;
  }
  long long p = pow(b * ny % MOD, k), ans = 0;
  if (p != 1)
    ans = (pow(p, (n + 1) / k) - 1) * pow(p - 1, MOD - 2) % MOD * lalal % MOD;
  else
    ans = lalal * ((n + 1) / k) % MOD;
  printf("%I64d", ans);
  return 0;
}

```