#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k;
char ch[100005];
long long mod = 1e9 + 9;
long long powd(long long x, long long y) {
  long long ans = 1;
  while (y > 0) {
    if (y % 2 == 1) ans = ans * x % mod;
    x = x * x % mod;
    y /= 2;
  }
  return (ans + mod) % mod;
}
long long afac(long long x) { return powd(x, mod - 2); }
int main() {
  while (~scanf("%lld%lld%lld%lld", &n, &a, &b, &k)) {
    scanf("%s", ch);
    long long ta = powd(a, n);
    long long tp = powd(b * afac(a) % mod, k);
    long long ans = 0;
    long long tn = (n + 1) / k;
    for (int i = 0; i < k; i++) {
      long long tans;
      if (tp != 1)
        tans = powd(b * afac(a) % mod, i) * ((1ll - powd(tp, tn)) % mod) % mod *
               afac(1ll - tp) % mod;
      else
        tans = powd(b * afac(a) % mod, i) * tn % mod;
      if (ch[i] == '-') tans = -tans;
      ans = (ans + tans % mod) % mod;
    }
    ans = (ans * ta) % mod;
    ans = (ans + mod) % mod;
    printf("%lld\n", ans);
  }
  return 0;
}

### 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>
using namespace std;
long long n, a, b, k;
char ch[100005];
long long mod = 1e9 + 9;
long long powd(long long x, long long y) {
  long long ans = 1;
  while (y > 0) {
    if (y % 2 == 1) ans = ans * x % mod;
    x = x * x % mod;
    y /= 2;
  }
  return (ans + mod) % mod;
}
long long afac(long long x) { return powd(x, mod - 2); }
int main() {
  while (~scanf("%lld%lld%lld%lld", &n, &a, &b, &k)) {
    scanf("%s", ch);
    long long ta = powd(a, n);
    long long tp = powd(b * afac(a) % mod, k);
    long long ans = 0;
    long long tn = (n + 1) / k;
    for (int i = 0; i < k; i++) {
      long long tans;
      if (tp != 1)
        tans = powd(b * afac(a) % mod, i) * ((1ll - powd(tp, tn)) % mod) % mod *
               afac(1ll - tp) % mod;
      else
        tans = powd(b * afac(a) % mod, i) * tn % mod;
      if (ch[i] == '-') tans = -tans;
      ans = (ans + tans % mod) % mod;
    }
    ans = (ans * ta) % mod;
    ans = (ans + mod) % mod;
    printf("%lld\n", ans);
  }
  return 0;
}

```