#include <bits/stdc++.h>
using namespace std;
int n, a, b, k, cnt;
long long ans, y[100005], mul, la;
char x[100005];
long long po(long long bb, int p) {
  long long re = 1;
  while (p) {
    if (p & 1) re = re * bb % 1000000009;
    bb = bb * bb % 1000000009;
    p >>= 1;
  }
  return re;
}
int main() {
  scanf("%d%d%d%d%s", &n, &a, &b, &k, x);
  y[0] = 1;
  for (int i = 1; i < k; i++) y[i] = y[i - 1] * b % 1000000009;
  la = po(a, n - (k - 1));
  for (int i = k - 1; i >= 0; i--) {
    if (x[i] == '+')
      ans = (ans + y[i] * la) % 1000000009;
    else
      ans = (ans + 1000000009 - y[i] * la % 1000000009) % 1000000009;
    la = la * a % 1000000009;
  }
  mul = po(b, k) * po(po(a, k), 1000000009 - 2) % 1000000009;
  cnt = (n + 1) / k;
  if (mul == 1)
    ans = ans * cnt % 1000000009;
  else
    ans = ans * (po(mul, cnt) + 1000000009 - 1) % 1000000009 *
          po((mul + 1000000009 - 1) % 1000000009, 1000000009 - 2) % 1000000009;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int n, a, b, k, cnt;
long long ans, y[100005], mul, la;
char x[100005];
long long po(long long bb, int p) {
  long long re = 1;
  while (p) {
    if (p & 1) re = re * bb % 1000000009;
    bb = bb * bb % 1000000009;
    p >>= 1;
  }
  return re;
}
int main() {
  scanf("%d%d%d%d%s", &n, &a, &b, &k, x);
  y[0] = 1;
  for (int i = 1; i < k; i++) y[i] = y[i - 1] * b % 1000000009;
  la = po(a, n - (k - 1));
  for (int i = k - 1; i >= 0; i--) {
    if (x[i] == '+')
      ans = (ans + y[i] * la) % 1000000009;
    else
      ans = (ans + 1000000009 - y[i] * la % 1000000009) % 1000000009;
    la = la * a % 1000000009;
  }
  mul = po(b, k) * po(po(a, k), 1000000009 - 2) % 1000000009;
  cnt = (n + 1) / k;
  if (mul == 1)
    ans = ans * cnt % 1000000009;
  else
    ans = ans * (po(mul, cnt) + 1000000009 - 1) % 1000000009 *
          po((mul + 1000000009 - 1) % 1000000009, 1000000009 - 2) % 1000000009;
  printf("%lld\n", ans);
  return 0;
}

```