#include <bits/stdc++.h>
using namespace std;
char aa[101010];
const int INF = 1000000009;
int ksm(int w, int x) {
  int ans = 1;
  while (x > 0) {
    if (x % 2 == 1) ans = (long long)ans * w % INF;
    x /= 2;
    w = (long long)w * w % INF;
  }
  return ans;
}
int main() {
  int i, n, a, b, k, times, tmpp, tmp, tmp1, ans;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  times = (n + 1) / k;
  scanf("%s", aa + 1);
  tmp = 0;
  for (i = 0; i < k; i++) {
    tmpp = (long long)ksm(a, n - i) * ksm(b, i) % INF;
    if (aa[i + 1] == '-') {
      tmp = (tmp - tmpp + INF) % INF;
    } else {
      tmp = (tmp + tmpp) % INF;
    }
  }
  tmpp = ksm(b, k);
  tmp1 = ksm(a, k);
  tmp1 = ksm(tmp1, INF - 2);
  tmpp = (long long)tmpp * tmp1 % INF;
  ans = 1;
  if (tmpp == 1)
    ans = times;
  else
    ans = (long long)(ksm(tmpp, times) - 1) * ksm(tmpp - 1, INF - 2) % INF;
  ans = (long long)ans * tmp % INF;
  printf("%d\n", ans);
  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;
char aa[101010];
const int INF = 1000000009;
int ksm(int w, int x) {
  int ans = 1;
  while (x > 0) {
    if (x % 2 == 1) ans = (long long)ans * w % INF;
    x /= 2;
    w = (long long)w * w % INF;
  }
  return ans;
}
int main() {
  int i, n, a, b, k, times, tmpp, tmp, tmp1, ans;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  times = (n + 1) / k;
  scanf("%s", aa + 1);
  tmp = 0;
  for (i = 0; i < k; i++) {
    tmpp = (long long)ksm(a, n - i) * ksm(b, i) % INF;
    if (aa[i + 1] == '-') {
      tmp = (tmp - tmpp + INF) % INF;
    } else {
      tmp = (tmp + tmpp) % INF;
    }
  }
  tmpp = ksm(b, k);
  tmp1 = ksm(a, k);
  tmp1 = ksm(tmp1, INF - 2);
  tmpp = (long long)tmpp * tmp1 % INF;
  ans = 1;
  if (tmpp == 1)
    ans = times;
  else
    ans = (long long)(ksm(tmpp, times) - 1) * ksm(tmpp - 1, INF - 2) % INF;
  ans = (long long)ans * tmp % INF;
  printf("%d\n", ans);
  return 0;
}

```