#include <bits/stdc++.h>
long long n, a, b, k;
char s;
const long long mod = 1000000009;
long long pow(long long x, long long y) {
  long long ans = 1;
  for (; y; y >>= 1, x = x * x % mod)
    if (y & 1) ans = ans * x % mod;
  return ans;
}
long long inv(long long x) {
  return x > 1 ? inv(mod % x) * (mod - mod / x) % mod : 1;
}
signed main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  a ^= b ^= a ^= b;
  long long t = a * inv(b) % mod;
  long long q = pow(t, k), p = pow(b, n), ans = 0, Q, QQ;
  long long y = (n + 1) / k;
  if (q != 1) Q = inv(q - 1), QQ = pow(q, y);
  for (long long i = 0; i < k; i++) {
    do s = getchar();
    while (s != '+' && s != '-');
    if (q == 1) {
      ans = (ans + (s == '+' ? 1 : -1) * p * y % mod + mod) % mod;
    } else {
      ans = (ans + (s == '+' ? 1 : -1) * p * (QQ - 1) % mod * Q % mod + mod) %
            mod;
    }
    p = p * t % mod;
  }
  printf("%d\n", 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>
long long n, a, b, k;
char s;
const long long mod = 1000000009;
long long pow(long long x, long long y) {
  long long ans = 1;
  for (; y; y >>= 1, x = x * x % mod)
    if (y & 1) ans = ans * x % mod;
  return ans;
}
long long inv(long long x) {
  return x > 1 ? inv(mod % x) * (mod - mod / x) % mod : 1;
}
signed main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  a ^= b ^= a ^= b;
  long long t = a * inv(b) % mod;
  long long q = pow(t, k), p = pow(b, n), ans = 0, Q, QQ;
  long long y = (n + 1) / k;
  if (q != 1) Q = inv(q - 1), QQ = pow(q, y);
  for (long long i = 0; i < k; i++) {
    do s = getchar();
    while (s != '+' && s != '-');
    if (q == 1) {
      ans = (ans + (s == '+' ? 1 : -1) * p * y % mod + mod) % mod;
    } else {
      ans = (ans + (s == '+' ? 1 : -1) * p * (QQ - 1) % mod * Q % mod + mod) %
            mod;
    }
    p = p * t % mod;
  }
  printf("%d\n", ans);
  return 0;
}

```