#include <bits/stdc++.h>
long long M = 1e9 + 9, n, a, b, k, ans, d, i;
char s[1 << 17];
inline long long P(long long a, long long b) {
  long long r = 1;
  do {
    if (b & 1) r = (r * a) % M;
    a = (a * a) % M;
  } while (b /= 2);
  return r;
}
int main() {
  std::cin >> n >> a >> b >> k >> s;
  d = P(b, k) * P(P(a, k), M - 2) % M;
  while (i < k) {
    long long a1 = P(a, n - i) * P(b, i) % M, l = (n - i) / k + 1,
              o = s[i] == '+' ? 1 : M - 1;
    if (P(d, l) != 1)
      ans += (o * (a1 * (P(d, l) % M + M - 1) % M) % M) * P(d - 1, M - 2) % M;
    else
      ans += (o * a1 % M) * l % M;
    ++i;
  }
  return std::cout << ans % M, 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
long long M = 1e9 + 9, n, a, b, k, ans, d, i;
char s[1 << 17];
inline long long P(long long a, long long b) {
  long long r = 1;
  do {
    if (b & 1) r = (r * a) % M;
    a = (a * a) % M;
  } while (b /= 2);
  return r;
}
int main() {
  std::cin >> n >> a >> b >> k >> s;
  d = P(b, k) * P(P(a, k), M - 2) % M;
  while (i < k) {
    long long a1 = P(a, n - i) * P(b, i) % M, l = (n - i) / k + 1,
              o = s[i] == '+' ? 1 : M - 1;
    if (P(d, l) != 1)
      ans += (o * (a1 * (P(d, l) % M + M - 1) % M) % M) * P(d - 1, M - 2) % M;
    else
      ans += (o * a1 % M) * l % M;
    ++i;
  }
  return std::cout << ans % M, 0;
}

```