#include <bits/stdc++.h>
const int MAXN = 1e5 + 10;
const long long MOD = 1e9 + 9;
long long n, a, b, k;
char opera[MAXN];
long long quick_pow(long long a, long long b, long long m) {
  if (!b) return 1 % m;
  if (b & 1) return a * quick_pow((a * a) % MOD, b >> 1, m) % m;
  return quick_pow((a * a) % MOD, b >> 1, m) % m;
}
long long divide_mod(long long a, long long b, long long m) {
  return (a % m) * quick_pow(b, m - 2, m) % m;
}
int main(void) {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", opera);
  long long q = divide_mod(quick_pow(b, k, MOD), quick_pow(a, k, MOD), MOD);
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long a0 = quick_pow(a, n - i, MOD) * quick_pow(b, i, MOD) % MOD;
    if (opera[i] == '-') a0 = -a0;
    long long sum = 0, m;
    m = (n - i) / k + 1;
    if (q == 1)
      sum = m * a0 % MOD;
    else
      sum = divide_mod(a0 * (1 - quick_pow(q, m, MOD)), (1 - q), MOD);
    ans = (ans + sum) % MOD;
  }
  if (ans < 0) ans += MOD;
  printf("%lld\n", 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>
const int MAXN = 1e5 + 10;
const long long MOD = 1e9 + 9;
long long n, a, b, k;
char opera[MAXN];
long long quick_pow(long long a, long long b, long long m) {
  if (!b) return 1 % m;
  if (b & 1) return a * quick_pow((a * a) % MOD, b >> 1, m) % m;
  return quick_pow((a * a) % MOD, b >> 1, m) % m;
}
long long divide_mod(long long a, long long b, long long m) {
  return (a % m) * quick_pow(b, m - 2, m) % m;
}
int main(void) {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", opera);
  long long q = divide_mod(quick_pow(b, k, MOD), quick_pow(a, k, MOD), MOD);
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long a0 = quick_pow(a, n - i, MOD) * quick_pow(b, i, MOD) % MOD;
    if (opera[i] == '-') a0 = -a0;
    long long sum = 0, m;
    m = (n - i) / k + 1;
    if (q == 1)
      sum = m * a0 % MOD;
    else
      sum = divide_mod(a0 * (1 - quick_pow(q, m, MOD)), (1 - q), MOD);
    ans = (ans + sum) % MOD;
  }
  if (ans < 0) ans += MOD;
  printf("%lld\n", ans);
  return 0;
}

```