#include <bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
const double eps = 1e-9;
const int maxn = 100000 + 10;
const long long mod = 1000000009;
long long M(long long a) { return (a % mod + mod) % mod; }
long long modpow(long long a, long long n) {
  long long b = 1;
  while (n) {
    if (n & 1) b = a * b % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long B = 1, A = modpow(a, n), IA = 1;
  long long ia = modpow(a, mod - 2);
  string s;
  cin >> s;
  long long x = 0;
  for (int i = 0; i < k; ++i) {
    long long v = s[i] == '+' ? +1 : mod - 1;
    x = (x + v * B % mod * A % mod) % mod;
    B = B * b % mod;
    A = A * ia % mod;
    IA = IA * ia % mod;
  }
  long long Y = B * IA % mod;
  long long t = (n + 1) / k;
  long long ans;
  if (Y == 1)
    ans = x * t % mod;
  else
    ans = x * (modpow(Y, t) - 1 + mod) % mod * modpow(Y - 1, mod - 2) % mod;
  cout << ans << '\n';
  return 0;
}

### Prompt

Please formulate 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 oo = 0x3f3f3f3f;
const double eps = 1e-9;
const int maxn = 100000 + 10;
const long long mod = 1000000009;
long long M(long long a) { return (a % mod + mod) % mod; }
long long modpow(long long a, long long n) {
  long long b = 1;
  while (n) {
    if (n & 1) b = a * b % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long B = 1, A = modpow(a, n), IA = 1;
  long long ia = modpow(a, mod - 2);
  string s;
  cin >> s;
  long long x = 0;
  for (int i = 0; i < k; ++i) {
    long long v = s[i] == '+' ? +1 : mod - 1;
    x = (x + v * B % mod * A % mod) % mod;
    B = B * b % mod;
    A = A * ia % mod;
    IA = IA * ia % mod;
  }
  long long Y = B * IA % mod;
  long long t = (n + 1) / k;
  long long ans;
  if (Y == 1)
    ans = x * t % mod;
  else
    ans = x * (modpow(Y, t) - 1 + mod) % mod * modpow(Y - 1, mod - 2) % mod;
  cout << ans << '\n';
  return 0;
}

```