#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k;
string s;
long long realMod(long long a, long long b) {
  a -= b;
  if (a < 0) a += 1000000009LL;
  return a;
}
long long binpow(long long x, long long n) {
  if (n == 0) return 1;
  if (n == 1) return x;
  long long t = binpow(x, n / 2);
  t = t * t % 1000000009LL;
  return n & 1 ? t * x % 1000000009LL : t;
}
signed main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long r =
      binpow(b, k) * binpow(binpow(a, k), 1000000009LL - 2) % 1000000009LL;
  long long ans = 0;
  for (long long i = 0; i < k; ++i) {
    long long lim = (n + 1) / k;
    long long init = binpow(a, n - i) * binpow(b, i) % 1000000009LL;
    if (s[i] == '-') init = 1000000009LL - init;
    if (r != 1)
      ans = (ans + init * realMod(1, binpow(r, lim)) % 1000000009LL *
                       binpow(realMod(1, r), 1000000009LL - 2)) %
            1000000009LL;
    else
      ans = (ans + init * lim) % 1000000009LL;
  }
  cout << ans;
  return 0;
}

### 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>
using namespace std;
long long n, a, b, k;
string s;
long long realMod(long long a, long long b) {
  a -= b;
  if (a < 0) a += 1000000009LL;
  return a;
}
long long binpow(long long x, long long n) {
  if (n == 0) return 1;
  if (n == 1) return x;
  long long t = binpow(x, n / 2);
  t = t * t % 1000000009LL;
  return n & 1 ? t * x % 1000000009LL : t;
}
signed main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long r =
      binpow(b, k) * binpow(binpow(a, k), 1000000009LL - 2) % 1000000009LL;
  long long ans = 0;
  for (long long i = 0; i < k; ++i) {
    long long lim = (n + 1) / k;
    long long init = binpow(a, n - i) * binpow(b, i) % 1000000009LL;
    if (s[i] == '-') init = 1000000009LL - init;
    if (r != 1)
      ans = (ans + init * realMod(1, binpow(r, lim)) % 1000000009LL *
                       binpow(realMod(1, r), 1000000009LL - 2)) %
            1000000009LL;
    else
      ans = (ans + init * lim) % 1000000009LL;
  }
  cout << ans;
  return 0;
}

```