#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using PII = pair<int, int>;
const int Mod = 1e9 + 9;
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  string s;
  cin >> s;
  auto pow = [&](int a, int k) {
    int ret = 1;
    int x = a;
    while (k) {
      if (k & 1) {
        ret = 1LL * ret * x % Mod;
      }
      x = 1LL * x * x % Mod;
      k >>= 1;
    }
    return ret;
  };
  function<int(int)> rev = [&](int a) {
    return a == 1 ? 1 : rev(Mod % a) * 1LL * (Mod - Mod / a) % Mod;
  };
  int ra = rev(a);
  int ini = 0;
  for (int t = 1LL * pow(a, n), i = 0; i <= k - 1; i++) {
    ini = (ini + Mod + (s[i] == '+' ? 1LL : -1LL) * t) % Mod;
    t = 1LL * t * ra % Mod * b % Mod;
  }
  int rak = rev(pow(a, k));
  int bk = pow(b, k);
  int x = (n + 1) / k;
  LL ans = 0;
  if ((1LL * bk * rak % Mod - 1 + Mod) % Mod == 0) {
    ans = 1LL * ini * x % Mod;
  } else
    ans = 1LL * ini * ((pow(1LL * bk * rak % Mod, x) - 1 + Mod) % Mod) % Mod *
          rev((1LL * bk * rak % Mod - 1 + Mod) % Mod) % Mod;
  printf("%lld\n", ans);
}

### Prompt

Please provide a cpp coded solution 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;
using LL = long long;
using PII = pair<int, int>;
const int Mod = 1e9 + 9;
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  string s;
  cin >> s;
  auto pow = [&](int a, int k) {
    int ret = 1;
    int x = a;
    while (k) {
      if (k & 1) {
        ret = 1LL * ret * x % Mod;
      }
      x = 1LL * x * x % Mod;
      k >>= 1;
    }
    return ret;
  };
  function<int(int)> rev = [&](int a) {
    return a == 1 ? 1 : rev(Mod % a) * 1LL * (Mod - Mod / a) % Mod;
  };
  int ra = rev(a);
  int ini = 0;
  for (int t = 1LL * pow(a, n), i = 0; i <= k - 1; i++) {
    ini = (ini + Mod + (s[i] == '+' ? 1LL : -1LL) * t) % Mod;
    t = 1LL * t * ra % Mod * b % Mod;
  }
  int rak = rev(pow(a, k));
  int bk = pow(b, k);
  int x = (n + 1) / k;
  LL ans = 0;
  if ((1LL * bk * rak % Mod - 1 + Mod) % Mod == 0) {
    ans = 1LL * ini * x % Mod;
  } else
    ans = 1LL * ini * ((pow(1LL * bk * rak % Mod, x) - 1 + Mod) % Mod) % Mod *
          rev((1LL * bk * rak % Mod - 1 + Mod) % Mod) % Mod;
  printf("%lld\n", ans);
}

```