#include <bits/stdc++.h>
using namespace std;
const int MX = 112345;
const long long mod = 1e9 + 9;
long long bigMod(long long b, long long p) {
  long long x = 1LL;
  for (; p > 0; p /= 2) {
    if (p & 1) x = (x * b) % mod;
    b = (b * b) % mod;
  }
  return x % mod;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, a, b, k, res = 0;
  string sgn;
  cin >> n >> a >> b >> k >> sgn;
  for (int i = 0; i < k; i++) {
    long long tmp = (bigMod(a, n - i) % mod * bigMod(b, i) % mod) % mod;
    if (sgn[i] == '-')
      res = (res - tmp + mod) % mod;
    else
      res = (res + tmp + mod) % mod;
  }
  long long p, q, d = ((n + 1) / k);
  p = (bigMod(b, k) % mod * bigMod(bigMod(a, k) % mod, mod - 2) % mod) % mod;
  if (p == 1) {
    res = (res * d + mod) % mod;
    cout << res << endl;
    return 0;
  }
  q = ((bigMod(p, d) - 1 + mod) % mod * bigMod(p - 1 + mod, mod - 2) % mod) %
      mod;
  res = (res * q + mod) % mod;
  cout << res << endl;
  return 0;
}

### Prompt

Generate 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 MX = 112345;
const long long mod = 1e9 + 9;
long long bigMod(long long b, long long p) {
  long long x = 1LL;
  for (; p > 0; p /= 2) {
    if (p & 1) x = (x * b) % mod;
    b = (b * b) % mod;
  }
  return x % mod;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, a, b, k, res = 0;
  string sgn;
  cin >> n >> a >> b >> k >> sgn;
  for (int i = 0; i < k; i++) {
    long long tmp = (bigMod(a, n - i) % mod * bigMod(b, i) % mod) % mod;
    if (sgn[i] == '-')
      res = (res - tmp + mod) % mod;
    else
      res = (res + tmp + mod) % mod;
  }
  long long p, q, d = ((n + 1) / k);
  p = (bigMod(b, k) % mod * bigMod(bigMod(a, k) % mod, mod - 2) % mod) % mod;
  if (p == 1) {
    res = (res * d + mod) % mod;
    cout << res << endl;
    return 0;
  }
  q = ((bigMod(p, d) - 1 + mod) % mod * bigMod(p - 1 + mod, mod - 2) % mod) %
      mod;
  res = (res * q + mod) % mod;
  cout << res << endl;
  return 0;
}

```