#include <bits/stdc++.h>
int n, A, B, K;
char s[100010];
long long ans;
long long INVA, INVB;
long long power(long long a, int b) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1) ret = ret * a % 1000000009;
    a = a * a % 1000000009;
  }
  return ret;
}
long long calcBase() {
  int i;
  long long x = power(A, n), y = 1, ret = 0;
  for (i = 0; i < K; i++) {
    ret = (ret + (s[i] == '+' ? 1 : -1) * x % 1000000009 * y % 1000000009 +
           1000000009) %
          1000000009;
    x = x * INVA % 1000000009;
    y = y * B % 1000000009;
  }
  return ret;
}
int main() {
  int i;
  long long x, q;
  scanf("%d%d%d%d%s", &n, &A, &B, &K, s);
  INVA = power(A, 1000000009 - 2);
  INVB = power(B, 1000000009 - 2);
  x = calcBase();
  q = power(INVA, K) * power(B, K) % 1000000009;
  if (q == 1)
    ans = x * ((n + 1) / K) % 1000000009;
  else {
    ans = x * (power(q, (n + 1) / K) - 1) % 1000000009;
    ans = ans * power(q - 1, 1000000009 - 2) % 1000000009;
  }
  printf("%I64d", ans);
  return 0;
}

### Prompt

Your task is to create 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>
int n, A, B, K;
char s[100010];
long long ans;
long long INVA, INVB;
long long power(long long a, int b) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1) ret = ret * a % 1000000009;
    a = a * a % 1000000009;
  }
  return ret;
}
long long calcBase() {
  int i;
  long long x = power(A, n), y = 1, ret = 0;
  for (i = 0; i < K; i++) {
    ret = (ret + (s[i] == '+' ? 1 : -1) * x % 1000000009 * y % 1000000009 +
           1000000009) %
          1000000009;
    x = x * INVA % 1000000009;
    y = y * B % 1000000009;
  }
  return ret;
}
int main() {
  int i;
  long long x, q;
  scanf("%d%d%d%d%s", &n, &A, &B, &K, s);
  INVA = power(A, 1000000009 - 2);
  INVB = power(B, 1000000009 - 2);
  x = calcBase();
  q = power(INVA, K) * power(B, K) % 1000000009;
  if (q == 1)
    ans = x * ((n + 1) / K) % 1000000009;
  else {
    ans = x * (power(q, (n + 1) / K) - 1) % 1000000009;
    ans = ans * power(q - 1, 1000000009 - 2) % 1000000009;
  }
  printf("%I64d", ans);
  return 0;
}

```