#include <bits/stdc++.h>
using namespace std;
const long long Base = 1e9 + 9;
long long Mul(long long a, long long b) { return (a * b + Base * Base) % Base; }
long long Add(long long a, long long b) { return (a + b + Base * Base) % Base; }
long long Pow(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long tmp = Pow(a, n / 2LL);
  tmp = Mul(tmp, tmp);
  if (n % 2LL) tmp = Mul(tmp, a);
  return tmp;
}
void load() {}
void process() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long ans = 0;
  for (long long i = 0; i < k; ++i) {
    char c;
    cin >> c;
    long long hs = 1;
    if (c == '-') hs = -1;
    long long res = hs;
    res = Mul(res, Pow(a, n - i));
    res = Mul(res, Pow(b, i));
    long long tmp = (n + 1LL) / k;
    long long tu = Pow(b, tmp * k);
    tu = Mul(tu, Pow(Pow(a, tmp * k), Base - 2LL));
    tu = Add(tu, -1LL);
    long long mau = Pow(b, k);
    mau = Mul(mau, Pow(Pow(a, k), Base - 2LL));
    mau = Add(mau, -1LL);
    long long sum = Mul(tu, Pow(mau, Base - 2LL));
    if (mau == 0) sum = tmp;
    ans = Add(ans, Mul(res, sum));
  }
  cout << ans % Base;
  return;
}
int main() {
  load();
  process();
  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 long long Base = 1e9 + 9;
long long Mul(long long a, long long b) { return (a * b + Base * Base) % Base; }
long long Add(long long a, long long b) { return (a + b + Base * Base) % Base; }
long long Pow(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a;
  long long tmp = Pow(a, n / 2LL);
  tmp = Mul(tmp, tmp);
  if (n % 2LL) tmp = Mul(tmp, a);
  return tmp;
}
void load() {}
void process() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long ans = 0;
  for (long long i = 0; i < k; ++i) {
    char c;
    cin >> c;
    long long hs = 1;
    if (c == '-') hs = -1;
    long long res = hs;
    res = Mul(res, Pow(a, n - i));
    res = Mul(res, Pow(b, i));
    long long tmp = (n + 1LL) / k;
    long long tu = Pow(b, tmp * k);
    tu = Mul(tu, Pow(Pow(a, tmp * k), Base - 2LL));
    tu = Add(tu, -1LL);
    long long mau = Pow(b, k);
    mau = Mul(mau, Pow(Pow(a, k), Base - 2LL));
    mau = Add(mau, -1LL);
    long long sum = Mul(tu, Pow(mau, Base - 2LL));
    if (mau == 0) sum = tmp;
    ans = Add(ans, Mul(res, sum));
  }
  cout << ans % Base;
  return;
}
int main() {
  load();
  process();
  return 0;
}

```