#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000009;
long long bigmod(long long n, long long po) {
  if (po == 0) return 1;
  if (po % 2 == 0) {
    long long ret = bigmod(n, po / 2);
    return (ret * ret) % mod;
  }
  long long ret = bigmod(n, po - 1);
  return (ret * n) % mod;
}
string SS;
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> SS;
  long long ans = 0;
  long long dddd = 0;
  for (int i = 0; i < SS.size(); i++) {
    long long N = n / k;
    if (N * k + i <= n) N++;
    long long A = bigmod(a, k);
    long long B = bigmod(b, k);
    long long up;
    up = ((bigmod(A, N) - bigmod(B, N)) % mod + mod) % mod;
    if (A == B) {
      long long zz = bigmod(b, i);
      long long zzz = bigmod(a, i);
      zzz = bigmod(zzz, mod - 2);
      zz = (zz * zzz) % mod;
      zz = (N * zz) % mod;
      up = (zz * bigmod(a, n)) % mod;
    } else {
      long long xx = bigmod(A, N);
      xx = bigmod(xx, mod - 2);
      up = (up * xx) % mod;
      long long Down = ((A - B) % mod + mod) % mod;
      long long d = bigmod(A, mod - 2);
      Down = (Down * d) % mod;
      Down = bigmod(Down, mod - 2);
      up = (up * Down) % mod;
      long long fst = bigmod(a, i);
      fst = bigmod(fst, mod - 2);
      long long Fst = bigmod(b, i);
      fst = (fst * Fst) % mod;
      up = (fst * up) % mod;
      long long zzzz = bigmod(a, n);
      up = (up * zzzz) % mod;
    }
    if (SS[i] == '+')
      ans = (ans + up) % mod;
    else {
      ans = ((ans - up) % mod + mod) % mod;
    }
  }
  cout << ans << endl;
}

### Prompt

In cpp, your task is to solve 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;
long long mod = 1000000009;
long long bigmod(long long n, long long po) {
  if (po == 0) return 1;
  if (po % 2 == 0) {
    long long ret = bigmod(n, po / 2);
    return (ret * ret) % mod;
  }
  long long ret = bigmod(n, po - 1);
  return (ret * n) % mod;
}
string SS;
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> SS;
  long long ans = 0;
  long long dddd = 0;
  for (int i = 0; i < SS.size(); i++) {
    long long N = n / k;
    if (N * k + i <= n) N++;
    long long A = bigmod(a, k);
    long long B = bigmod(b, k);
    long long up;
    up = ((bigmod(A, N) - bigmod(B, N)) % mod + mod) % mod;
    if (A == B) {
      long long zz = bigmod(b, i);
      long long zzz = bigmod(a, i);
      zzz = bigmod(zzz, mod - 2);
      zz = (zz * zzz) % mod;
      zz = (N * zz) % mod;
      up = (zz * bigmod(a, n)) % mod;
    } else {
      long long xx = bigmod(A, N);
      xx = bigmod(xx, mod - 2);
      up = (up * xx) % mod;
      long long Down = ((A - B) % mod + mod) % mod;
      long long d = bigmod(A, mod - 2);
      Down = (Down * d) % mod;
      Down = bigmod(Down, mod - 2);
      up = (up * Down) % mod;
      long long fst = bigmod(a, i);
      fst = bigmod(fst, mod - 2);
      long long Fst = bigmod(b, i);
      fst = (fst * Fst) % mod;
      up = (fst * up) % mod;
      long long zzzz = bigmod(a, n);
      up = (up * zzzz) % mod;
    }
    if (SS[i] == '+')
      ans = (ans + up) % mod;
    else {
      ans = ((ans - up) % mod + mod) % mod;
    }
  }
  cout << ans << endl;
}

```