#include <bits/stdc++.h>
long long pow(long long a, long long n, long long mod) {
  long long t;
  for (t = 1; n; n >>= 1, a *= a, a %= mod) n & 1 && (t *= a, t %= mod);
  return t;
}
int main() {
  long long N, A, B, K, i, l, t, ans, mod = 1000000009;
  char s[100100];
  for (scanf("%lld%lld%lld%lld%s", &N, &A, &B, &K, s), ans = i = 0;
       i <= N && s[i]; i++)
    s[i] == '+' ? ans += pow(A, N - i, mod) * pow(B, i, mod)
                : (ans -= pow(A, N - i, mod) * pow(B, i, mod)),
        ans %= mod;
  if (i <= N)
    for (l = i, t = pow(pow(A, i, mod), mod - 2, mod) * pow(B, i, mod) % mod,
        ans = t - 1 % mod ? (ans * pow(t, (N + 1) / i, mod) - ans) % mod *
                                pow(t - 1, mod - 2, mod) % mod
                          : (N + 1) / i * ans % mod,
        i += (N + 1) / i * i;
         i <= N; i++)
      s[i % l] == '+' ? ans += pow(A, N - i, mod) * pow(B, i, mod)
                      : (ans -= pow(A, N - i, mod) * pow(B, i, mod)),
          ans %= mod;
  printf("%lld\n", (ans + mod) % mod);
}

### Prompt

Your challenge is to write 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 pow(long long a, long long n, long long mod) {
  long long t;
  for (t = 1; n; n >>= 1, a *= a, a %= mod) n & 1 && (t *= a, t %= mod);
  return t;
}
int main() {
  long long N, A, B, K, i, l, t, ans, mod = 1000000009;
  char s[100100];
  for (scanf("%lld%lld%lld%lld%s", &N, &A, &B, &K, s), ans = i = 0;
       i <= N && s[i]; i++)
    s[i] == '+' ? ans += pow(A, N - i, mod) * pow(B, i, mod)
                : (ans -= pow(A, N - i, mod) * pow(B, i, mod)),
        ans %= mod;
  if (i <= N)
    for (l = i, t = pow(pow(A, i, mod), mod - 2, mod) * pow(B, i, mod) % mod,
        ans = t - 1 % mod ? (ans * pow(t, (N + 1) / i, mod) - ans) % mod *
                                pow(t - 1, mod - 2, mod) % mod
                          : (N + 1) / i * ans % mod,
        i += (N + 1) / i * i;
         i <= N; i++)
      s[i % l] == '+' ? ans += pow(A, N - i, mod) * pow(B, i, mod)
                      : (ans -= pow(A, N - i, mod) * pow(B, i, mod)),
          ans %= mod;
  printf("%lld\n", (ans + mod) % mod);
}

```