#include <bits/stdc++.h>
using namespace std;
inline void setmin(int &x, int y) {
  if (y < x) x = y;
}
inline void setmax(int &x, int y) {
  if (y > x) x = y;
}
inline void setmin(long long &x, long long y) {
  if (y < x) x = y;
}
inline void setmax(long long &x, long long y) {
  if (y > x) x = y;
}
const int N = 100000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 9;
const int MOD1 = (int)1e9 + 9;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
long long bin_pow(long long a, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * a % MOD;
    a = a * a % MOD;
    p >>= 1;
  }
  return res;
}
long long rev_mod(long long a) { return bin_pow(a, MOD - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long ans = 0;
  long long cnt = (n + 1) / k;
  long long BA = b * rev_mod(a) % MOD;
  long long cur;
  if (bin_pow(BA, k) == 1)
    cur = bin_pow(a, n) * cnt % MOD;
  else
    cur = bin_pow(a, n) * (bin_pow(bin_pow(BA, k), cnt) - 1 + MOD) % MOD *
          rev_mod((bin_pow(BA, k) - 1 + MOD) % MOD) % MOD;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      ans = (ans + cur) % MOD;
    else
      ans = (ans - cur + MOD) % MOD;
    cur = cur * BA % MOD;
  }
  cout << ans << "\n";
  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;
inline void setmin(int &x, int y) {
  if (y < x) x = y;
}
inline void setmax(int &x, int y) {
  if (y > x) x = y;
}
inline void setmin(long long &x, long long y) {
  if (y < x) x = y;
}
inline void setmax(long long &x, long long y) {
  if (y > x) x = y;
}
const int N = 100000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 9;
const int MOD1 = (int)1e9 + 9;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
long long bin_pow(long long a, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * a % MOD;
    a = a * a % MOD;
    p >>= 1;
  }
  return res;
}
long long rev_mod(long long a) { return bin_pow(a, MOD - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  string s;
  cin >> s;
  long long ans = 0;
  long long cnt = (n + 1) / k;
  long long BA = b * rev_mod(a) % MOD;
  long long cur;
  if (bin_pow(BA, k) == 1)
    cur = bin_pow(a, n) * cnt % MOD;
  else
    cur = bin_pow(a, n) * (bin_pow(bin_pow(BA, k), cnt) - 1 + MOD) % MOD *
          rev_mod((bin_pow(BA, k) - 1 + MOD) % MOD) % MOD;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      ans = (ans + cur) % MOD;
    else
      ans = (ans - cur + MOD) % MOD;
    cur = cur * BA % MOD;
  }
  cout << ans << "\n";
  return 0;
}

```