#include <bits/stdc++.h>
using namespace std;
int vv[65];
inline int ksm(int p, int k) {
  vv[0] = p;
  for (int i = 1; i <= 29; ++i)
    vv[i] = (long long)vv[i - 1] * vv[i - 1] % 1000000009;
  int s = 1;
  for (int i = 29; i >= 0; --i) {
    if (1 << i <= k) {
      k -= 1 << i;
      s = (long long)s * vv[i] % 1000000009;
    }
  }
  return s;
}
long long ll, a_, b_, s, l, ans, q, n, a, b, k;
char ss[1000050];
long long aa[1000050];
inline void upt(long long &a, long long b) { a = (a + b) % 1000000009; }
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", ss + 1);
  for (int i = 1; i <= k; ++i)
    if (ss[i] == '+')
      aa[i - 1] = 1;
    else
      aa[i - 1] = 1000000009 - 1;
  for (int i = 0; i < k; ++i)
    upt(s, aa[i] * ksm(a, n - i) % 1000000009 * ksm(b, i) % 1000000009);
  a_ = ksm(a, k);
  b_ = ksm(b, k);
  q = b_ * ksm(a_, 1000000009 - 2) % 1000000009;
  if (q == 1) {
    l = (n + 1) / k;
    s = s * l % 1000000009;
    cout << s;
    return 0;
  }
  l = (n + 1) / k;
  ans = (ksm(q, l) + 1000000009 - 1) % 1000000009;
  ans = ans * ksm(q + 1000000009 - 1, 1000000009 - 2) % 1000000009;
  ans = ans * s % 1000000009;
  cout << ans;
}

### Prompt

Please create a solution in Cpp 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;
int vv[65];
inline int ksm(int p, int k) {
  vv[0] = p;
  for (int i = 1; i <= 29; ++i)
    vv[i] = (long long)vv[i - 1] * vv[i - 1] % 1000000009;
  int s = 1;
  for (int i = 29; i >= 0; --i) {
    if (1 << i <= k) {
      k -= 1 << i;
      s = (long long)s * vv[i] % 1000000009;
    }
  }
  return s;
}
long long ll, a_, b_, s, l, ans, q, n, a, b, k;
char ss[1000050];
long long aa[1000050];
inline void upt(long long &a, long long b) { a = (a + b) % 1000000009; }
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", ss + 1);
  for (int i = 1; i <= k; ++i)
    if (ss[i] == '+')
      aa[i - 1] = 1;
    else
      aa[i - 1] = 1000000009 - 1;
  for (int i = 0; i < k; ++i)
    upt(s, aa[i] * ksm(a, n - i) % 1000000009 * ksm(b, i) % 1000000009);
  a_ = ksm(a, k);
  b_ = ksm(b, k);
  q = b_ * ksm(a_, 1000000009 - 2) % 1000000009;
  if (q == 1) {
    l = (n + 1) / k;
    s = s * l % 1000000009;
    cout << s;
    return 0;
  }
  l = (n + 1) / k;
  ans = (ksm(q, l) + 1000000009 - 1) % 1000000009;
  ans = ans * ksm(q + 1000000009 - 1, 1000000009 - 2) % 1000000009;
  ans = ans * s % 1000000009;
  cout << ans;
}

```