#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 9;
const int maxn = 1e5 + 45;
char s[maxn];
long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans = ans * a % mod;
    }
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  int len = strlen(s);
  long long inv = qpow(b * qpow(a, mod - 2) % mod, k);
  long long ans = 0;
  for (int i = 0; i < len; i++) {
    long long tp;
    if (inv == 1) {
      tp = qpow(a, n - i) % mod * qpow(b, i) % mod;
      long long p = (n - i) / k;
      tp = tp * (p + 1) % mod;
    } else {
      tp = qpow(a, n - i) % mod * qpow(b, i) % mod;
      long long p = (n - i) / k;
      tp = tp * ((1 - qpow(inv, p + 1) + mod) % mod) % mod *
           qpow((1 - inv + mod) % mod, mod - 2) % mod;
    }
    if (s[i] == '+') {
      ans += tp;
    } else {
      ans -= tp;
    }
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans;
}

### Prompt

Your challenge is to write 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;
long long mod = 1e9 + 9;
const int maxn = 1e5 + 45;
char s[maxn];
long long qpow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) {
      ans = ans * a % mod;
    }
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  int len = strlen(s);
  long long inv = qpow(b * qpow(a, mod - 2) % mod, k);
  long long ans = 0;
  for (int i = 0; i < len; i++) {
    long long tp;
    if (inv == 1) {
      tp = qpow(a, n - i) % mod * qpow(b, i) % mod;
      long long p = (n - i) / k;
      tp = tp * (p + 1) % mod;
    } else {
      tp = qpow(a, n - i) % mod * qpow(b, i) % mod;
      long long p = (n - i) / k;
      tp = tp * ((1 - qpow(inv, p + 1) + mod) % mod) % mod *
           qpow((1 - inv + mod) % mod, mod - 2) % mod;
    }
    if (s[i] == '+') {
      ans += tp;
    } else {
      ans -= tp;
    }
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans;
}

```