#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k, a1;
char s[600005];
inline long long ksm(long long x, long long y) {
  long long ans1 = 1;
  while (y) {
    if (y & 1) ans1 = 1ll * ans1 * x % 1000000009;
    y >>= 1;
    x = 1ll * x * x % 1000000009;
  }
  return ans1;
}
signed main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long opt = 0;
  for (long long i = 0; i < k; i++) {
    if (s[i] == '-')
      opt = -1;
    else
      opt = 1;
    a1 = 1ll * (a1 + 1ll * opt * ksm(a, n - i) * ksm(b, i) % 1000000009) %
         1000000009;
    a1 = (a1 + 1000000009) % 1000000009;
  }
  long long xi = n + 1;
  long long num = xi / k;
  xi = xi - num * k;
  long long tmpa = ksm(a, k * num);
  long long tmpb = ksm(b, k * num);
  if (tmpa == tmpb) {
    printf("%lld\n", 1ll * a1 * num % 1000000009);
    return 0;
  }
  long long ans1 =
      1ll * (1ll * a1 * tmpa % 1000000009 - 1ll * a1 * tmpb % 1000000009) %
      1000000009;
  ans1 = (ans1 + 1000000009) % 1000000009;
  long long fm =
      (tmpa - 1ll * ksm(b, k) * ksm(a, k * num - k) % 1000000009) % 1000000009;
  fm = (fm + 1000000009) % 1000000009;
  ans1 = 1ll * ans1 * ksm(fm, 1000000009 - 2) % 1000000009;
  printf("%lld\n", ans1 % 1000000009);
  return 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>
using namespace std;
long long n, a, b, k, a1;
char s[600005];
inline long long ksm(long long x, long long y) {
  long long ans1 = 1;
  while (y) {
    if (y & 1) ans1 = 1ll * ans1 * x % 1000000009;
    y >>= 1;
    x = 1ll * x * x % 1000000009;
  }
  return ans1;
}
signed main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long opt = 0;
  for (long long i = 0; i < k; i++) {
    if (s[i] == '-')
      opt = -1;
    else
      opt = 1;
    a1 = 1ll * (a1 + 1ll * opt * ksm(a, n - i) * ksm(b, i) % 1000000009) %
         1000000009;
    a1 = (a1 + 1000000009) % 1000000009;
  }
  long long xi = n + 1;
  long long num = xi / k;
  xi = xi - num * k;
  long long tmpa = ksm(a, k * num);
  long long tmpb = ksm(b, k * num);
  if (tmpa == tmpb) {
    printf("%lld\n", 1ll * a1 * num % 1000000009);
    return 0;
  }
  long long ans1 =
      1ll * (1ll * a1 * tmpa % 1000000009 - 1ll * a1 * tmpb % 1000000009) %
      1000000009;
  ans1 = (ans1 + 1000000009) % 1000000009;
  long long fm =
      (tmpa - 1ll * ksm(b, k) * ksm(a, k * num - k) % 1000000009) % 1000000009;
  fm = (fm + 1000000009) % 1000000009;
  ans1 = 1ll * ans1 * ksm(fm, 1000000009 - 2) % 1000000009;
  printf("%lld\n", ans1 % 1000000009);
  return 0;
}

```