#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long M = 1e9 + 9;
const long long MAXN = 1e5 + 10;
long long n, k, a, b, s[MAXN];
long long bpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = (res * a) % M;
    a = (a * a) % M;
    b /= 2;
  }
  return res;
}
void add(long long &a, long long b) { a = (a + b) % M; }
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> a >> b >> k;
  long long total = 0;
  for (long long i = 0; i < k; i++) {
    char c;
    cin >> c;
    s[i] = (c == '+' ? 1 : M - 1);
  }
  for (long long i = 0; i < k; i++) {
    add(total, s[i] * (bpow(a, n - i) * bpow(b, i) % M) % M);
  }
  long long q = bpow(b, k) * bpow(bpow(a, k), M - 2) % M;
  long long ans = total *
                  (q == 1 ? (n + 1) / k
                          : (bpow(q, (n + 1) / k) + M - 1) % M *
                                bpow(q + M - 1, M - 2) % M) %
                  M;
  cout << ans % M;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const long long M = 1e9 + 9;
const long long MAXN = 1e5 + 10;
long long n, k, a, b, s[MAXN];
long long bpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = (res * a) % M;
    a = (a * a) % M;
    b /= 2;
  }
  return res;
}
void add(long long &a, long long b) { a = (a + b) % M; }
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> a >> b >> k;
  long long total = 0;
  for (long long i = 0; i < k; i++) {
    char c;
    cin >> c;
    s[i] = (c == '+' ? 1 : M - 1);
  }
  for (long long i = 0; i < k; i++) {
    add(total, s[i] * (bpow(a, n - i) * bpow(b, i) % M) % M);
  }
  long long q = bpow(b, k) * bpow(bpow(a, k), M - 2) % M;
  long long ans = total *
                  (q == 1 ? (n + 1) / k
                          : (bpow(q, (n + 1) / k) + M - 1) % M *
                                bpow(q + M - 1, M - 2) % M) %
                  M;
  cout << ans % M;
}

```