#include <bits/stdc++.h>
using namespace std;
constexpr int N = 1e6 + 10;
constexpr int MOD = 1e9 + 9;
long long n, a, b, k, sum;
string s;
long long mypow(long long a, long long b) {
  return b ? (mypow(a * a % MOD, b >> 1) * (b & 1 ? a : 1)) % MOD : 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> a >> b >> k >> s;
  for (int i = 0; i < k; i++) {
    long long local = mypow(a, n - i) * mypow(b, i);
    local %= MOD;
    if (s[i] == '-')
      sum -= local;
    else
      sum += local;
    sum = ((sum % MOD) + MOD) % MOD;
  }
  long long tedad = (n + 1) / k;
  if (a == b) {
    long long ans = sum * tedad;
    ans %= MOD;
    if (ans < 0) ans += MOD;
    cout << ans << '\n';
    return 0;
  }
  long long r = mypow(b, k) * mypow(a, MOD - 1 - k) % MOD;
  if (r == 1) {
    long long ans = sum * tedad;
    ans %= MOD;
    if (ans < 0) ans += MOD;
    cout << ans << '\n';
    return 0;
  }
  long long ans = 1ll - mypow(r, tedad);
  ans = ((ans % MOD) + MOD) % MOD;
  ans = ans * mypow((1 - r) % MOD, MOD - 2);
  ans %= MOD;
  if (ans < 0) ans += MOD;
  ans *= sum;
  ans %= MOD;
  if (ans < 0) ans += MOD;
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in cpp 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;
constexpr int N = 1e6 + 10;
constexpr int MOD = 1e9 + 9;
long long n, a, b, k, sum;
string s;
long long mypow(long long a, long long b) {
  return b ? (mypow(a * a % MOD, b >> 1) * (b & 1 ? a : 1)) % MOD : 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> a >> b >> k >> s;
  for (int i = 0; i < k; i++) {
    long long local = mypow(a, n - i) * mypow(b, i);
    local %= MOD;
    if (s[i] == '-')
      sum -= local;
    else
      sum += local;
    sum = ((sum % MOD) + MOD) % MOD;
  }
  long long tedad = (n + 1) / k;
  if (a == b) {
    long long ans = sum * tedad;
    ans %= MOD;
    if (ans < 0) ans += MOD;
    cout << ans << '\n';
    return 0;
  }
  long long r = mypow(b, k) * mypow(a, MOD - 1 - k) % MOD;
  if (r == 1) {
    long long ans = sum * tedad;
    ans %= MOD;
    if (ans < 0) ans += MOD;
    cout << ans << '\n';
    return 0;
  }
  long long ans = 1ll - mypow(r, tedad);
  ans = ((ans % MOD) + MOD) % MOD;
  ans = ans * mypow((1 - r) % MOD, MOD - 2);
  ans %= MOD;
  if (ans < 0) ans += MOD;
  ans *= sum;
  ans %= MOD;
  if (ans < 0) ans += MOD;
  cout << ans << '\n';
  return 0;
}

```