#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k;
long long mod = (long long)1e9 + 9;
long long ases[(long long)1e5 + 5];
long long bs[(long long)1e5 + 5];
long long fas_expo(long long base, long long pw) {
  long long ans = 1;
  while (pw > 0) {
    if (pw % 2 == 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    pw /= 2;
  }
  return ans;
}
long long mod_inv(long long a2) { return fas_expo(a2, mod - 2); }
void cons() {
  ases[0] = fas_expo(a, n);
  bs[0] = 1;
  for (int i = 1; i <= k; i++) {
    ases[i] = ases[i - 1] * mod_inv(a) % mod;
    bs[i] = bs[i - 1] * b % mod;
  }
}
int main() {
  cin >> n >> a >> b >> k;
  cons();
  string s;
  cin >> s;
  long long x = (n + 1) / k - 1;
  long long h = (b * mod_inv(a)) % mod;
  h = fas_expo(h, k);
  long long d = fas_expo(h, x + 1) - 1;
  d = d * mod_inv(h - 1) % mod;
  if (h == 1) {
    d = x + 1;
  }
  long long t = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      t = (t + ases[i] * bs[i]) % mod;
    } else {
      t = (t - ases[i] * bs[i]) % mod;
    }
  }
  long long res = t * d % mod;
  if (res < 0) {
    res += mod;
  }
  cout << res << endl;
  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;
long long n, a, b, k;
long long mod = (long long)1e9 + 9;
long long ases[(long long)1e5 + 5];
long long bs[(long long)1e5 + 5];
long long fas_expo(long long base, long long pw) {
  long long ans = 1;
  while (pw > 0) {
    if (pw % 2 == 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    pw /= 2;
  }
  return ans;
}
long long mod_inv(long long a2) { return fas_expo(a2, mod - 2); }
void cons() {
  ases[0] = fas_expo(a, n);
  bs[0] = 1;
  for (int i = 1; i <= k; i++) {
    ases[i] = ases[i - 1] * mod_inv(a) % mod;
    bs[i] = bs[i - 1] * b % mod;
  }
}
int main() {
  cin >> n >> a >> b >> k;
  cons();
  string s;
  cin >> s;
  long long x = (n + 1) / k - 1;
  long long h = (b * mod_inv(a)) % mod;
  h = fas_expo(h, k);
  long long d = fas_expo(h, x + 1) - 1;
  d = d * mod_inv(h - 1) % mod;
  if (h == 1) {
    d = x + 1;
  }
  long long t = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      t = (t + ases[i] * bs[i]) % mod;
    } else {
      t = (t - ases[i] * bs[i]) % mod;
    }
  }
  long long res = t * d % mod;
  if (res < 0) {
    res += mod;
  }
  cout << res << endl;
  return 0;
}

```