#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k;
char buf[100010];
long long qp(long long x, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = ret * x % 1000000009;
    x = x * x % 1000000009;
    p >>= 1;
  }
  return ret;
}
long long rev(long long x) { return qp(x, 1000000009 - 2); }
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", buf + 1);
  long long num = qp(a, n);
  long long q = b * rev(a) % 1000000009;
  long long qq = qp(q, k);
  long long tmp = (n + 1) / k;
  long long ans = 0;
  for (long long i = 1; i <= k; i++) {
    long long cur = num;
    if (qq != 1) {
      cur = cur * rev((qq - 1 + 1000000009) % 1000000009) % 1000000009;
      cur = cur * ((qp(qq, tmp) - 1 + 1000000009) % 1000000009) % 1000000009;
    } else {
      cur = num * tmp % 1000000009;
    }
    if (buf[i] == '-')
      ans = (ans - cur + 1000000009) % 1000000009;
    else
      ans = (ans + cur) % 1000000009;
    num = num * q % 1000000009;
  }
  printf("%lld\n", ans);
}

### Prompt

Please create a solution in Cpp 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 buf[100010];
long long qp(long long x, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = ret * x % 1000000009;
    x = x * x % 1000000009;
    p >>= 1;
  }
  return ret;
}
long long rev(long long x) { return qp(x, 1000000009 - 2); }
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", buf + 1);
  long long num = qp(a, n);
  long long q = b * rev(a) % 1000000009;
  long long qq = qp(q, k);
  long long tmp = (n + 1) / k;
  long long ans = 0;
  for (long long i = 1; i <= k; i++) {
    long long cur = num;
    if (qq != 1) {
      cur = cur * rev((qq - 1 + 1000000009) % 1000000009) % 1000000009;
      cur = cur * ((qp(qq, tmp) - 1 + 1000000009) % 1000000009) % 1000000009;
    } else {
      cur = num * tmp % 1000000009;
    }
    if (buf[i] == '-')
      ans = (ans - cur + 1000000009) % 1000000009;
    else
      ans = (ans + cur) % 1000000009;
    num = num * q % 1000000009;
  }
  printf("%lld\n", ans);
}

```