#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int maxint = 0x3f3f3f3f;
const int mod = 1e9 + 9;
long long n, a, b, k, f[maxn], ans;
char c;
long long qp(long long x, long long y) {
  if (y == 0) return (long long)1;
  long long z = qp(x, y / 2);
  if (y % 2 == 0)
    return z * z % mod;
  else
    return z * z % mod * x % mod;
}
int main() {
  long long i, j, h, g, p;
  cin >> n >> a >> b >> k;
  n++;
  for (i = 1; i <= k; i++) {
    cin >> c;
    if (c == '+')
      f[i] = (long long)1;
    else
      f[i] = (long long)-1;
  }
  for (i = 1; i <= min(n, k); i++) {
    j = (n - i) / k + 1;
    h = f[i] * qp(a, n - i) * qp(b, i - 1) % mod;
    g = qp(qp(a, mod - 2), k) * qp(b, k) % mod;
    if (g == 1)
      ans = (ans + h * j % mod) % mod;
    else {
      p = qp(g, j);
      ans = (ans + h * (p - 1) % mod * qp(g - 1, mod - 2) % mod) % mod;
    }
  }
  if (ans < 0) ans += mod;
  cout << ans << "\n";
  return 0;
}

### Prompt

Your task is to create 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;
const int maxn = 1e5 + 10;
const int maxint = 0x3f3f3f3f;
const int mod = 1e9 + 9;
long long n, a, b, k, f[maxn], ans;
char c;
long long qp(long long x, long long y) {
  if (y == 0) return (long long)1;
  long long z = qp(x, y / 2);
  if (y % 2 == 0)
    return z * z % mod;
  else
    return z * z % mod * x % mod;
}
int main() {
  long long i, j, h, g, p;
  cin >> n >> a >> b >> k;
  n++;
  for (i = 1; i <= k; i++) {
    cin >> c;
    if (c == '+')
      f[i] = (long long)1;
    else
      f[i] = (long long)-1;
  }
  for (i = 1; i <= min(n, k); i++) {
    j = (n - i) / k + 1;
    h = f[i] * qp(a, n - i) * qp(b, i - 1) % mod;
    g = qp(qp(a, mod - 2), k) * qp(b, k) % mod;
    if (g == 1)
      ans = (ans + h * j % mod) % mod;
    else {
      p = qp(g, j);
      ans = (ans + h * (p - 1) % mod * qp(g - 1, mod - 2) % mod) % mod;
    }
  }
  if (ans < 0) ans += mod;
  cout << ans << "\n";
  return 0;
}

```