#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long n, a, b, k, mod = 1e9 + 9, t, s;
char g;
long long pwr(long long a, long long p) {
  if (!p) return 1;
  long long t = pwr(a, p / 2);
  if (p % 2)
    return t * t % mod * a % mod;
  else
    return t * t % mod;
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n >> a >> b >> k;
  for (int i = 1; i <= k; i++) {
    cin >> g;
    if (g == '+')
      s = 1;
    else
      s = -1;
    t += s * pwr(a, n - i + 1) % mod * pwr(b, i - 1) % mod;
    t %= mod;
    t += mod;
    t %= mod;
  }
  long long f = pwr(b, k) * pwr(pwr(a, k), mod - 2) % mod;
  if (f != 1) {
    long long p = (n + 1) / k - 1;
    t *= (pwr(f, p + 1) - 1 + mod) % mod;
    t %= mod;
    t *= pwr(f - 1, mod - 2);
    t %= mod;
  } else {
    long long p = (n + 1) / k;
    t *= p;
    t %= mod;
  }
  cout << t;
}

### 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>
#pragma GCC optimize("O3")
using namespace std;
long long n, a, b, k, mod = 1e9 + 9, t, s;
char g;
long long pwr(long long a, long long p) {
  if (!p) return 1;
  long long t = pwr(a, p / 2);
  if (p % 2)
    return t * t % mod * a % mod;
  else
    return t * t % mod;
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n >> a >> b >> k;
  for (int i = 1; i <= k; i++) {
    cin >> g;
    if (g == '+')
      s = 1;
    else
      s = -1;
    t += s * pwr(a, n - i + 1) % mod * pwr(b, i - 1) % mod;
    t %= mod;
    t += mod;
    t %= mod;
  }
  long long f = pwr(b, k) * pwr(pwr(a, k), mod - 2) % mod;
  if (f != 1) {
    long long p = (n + 1) / k - 1;
    t *= (pwr(f, p + 1) - 1 + mod) % mod;
    t %= mod;
    t *= pwr(f - 1, mod - 2);
    t %= mod;
  } else {
    long long p = (n + 1) / k;
    t *= p;
    t %= mod;
  }
  cout << t;
}

```