#include <bits/stdc++.h>
using namespace std;
template <class BidirIt>
BidirIt prev(BidirIt it,
             typename iterator_traits<BidirIt>::difference_type n = 1) {
  advance(it, -n);
  return it;
}
template <class ForwardIt>
ForwardIt next(ForwardIt it,
               typename iterator_traits<ForwardIt>::difference_type n = 1) {
  advance(it, n);
  return it;
}
const double EPS = 1e-9;
const double PI = 3.141592653589793238462;
template <typename T>
inline T sq(T a) {
  return a * a;
}
const int MOD = 1e9 + 9;
inline int mul_mod(int a, int b) { return a * 1LL * b % MOD; }
inline void add_mod(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
inline void sub_mod(int &a, int b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mpow(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mul_mod(res, a);
    a = mul_mod(a, a);
    b >>= 1;
  }
  return res;
}
int calc(int a, int b) {
  if (b == 0) return 1;
  int res = calc(a, b / 2);
  int add = res;
  if (b % 2 == 0) sub_mod(add, mpow(a, b / 2));
  add_mod(res, mul_mod(mpow(a, b / 2 + 1), add));
  return res;
}
int main() {
  int n, a, b, k;
  scanf("%d %d %d %d ", &n, &a, &b, &k);
  int ans = 0;
  for (int i = 0; i < k; i++) {
    if (getchar_unlocked() == '+')
      add_mod(ans, mul_mod(mpow(b, i), mpow(a, n - i)));
    else
      sub_mod(ans, mul_mod(mpow(b, i), mpow(a, n - i)));
  }
  int c = mul_mod(mpow(mpow(a, k), MOD - 2), mpow(b, k));
  printf("%d\n", mul_mod(ans, calc(c, (n + 1) / k - 1)));
  return 0;
}

### Prompt

Create a solution in CPP for 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;
template <class BidirIt>
BidirIt prev(BidirIt it,
             typename iterator_traits<BidirIt>::difference_type n = 1) {
  advance(it, -n);
  return it;
}
template <class ForwardIt>
ForwardIt next(ForwardIt it,
               typename iterator_traits<ForwardIt>::difference_type n = 1) {
  advance(it, n);
  return it;
}
const double EPS = 1e-9;
const double PI = 3.141592653589793238462;
template <typename T>
inline T sq(T a) {
  return a * a;
}
const int MOD = 1e9 + 9;
inline int mul_mod(int a, int b) { return a * 1LL * b % MOD; }
inline void add_mod(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
inline void sub_mod(int &a, int b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mpow(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mul_mod(res, a);
    a = mul_mod(a, a);
    b >>= 1;
  }
  return res;
}
int calc(int a, int b) {
  if (b == 0) return 1;
  int res = calc(a, b / 2);
  int add = res;
  if (b % 2 == 0) sub_mod(add, mpow(a, b / 2));
  add_mod(res, mul_mod(mpow(a, b / 2 + 1), add));
  return res;
}
int main() {
  int n, a, b, k;
  scanf("%d %d %d %d ", &n, &a, &b, &k);
  int ans = 0;
  for (int i = 0; i < k; i++) {
    if (getchar_unlocked() == '+')
      add_mod(ans, mul_mod(mpow(b, i), mpow(a, n - i)));
    else
      sub_mod(ans, mul_mod(mpow(b, i), mpow(a, n - i)));
  }
  int c = mul_mod(mpow(mpow(a, k), MOD - 2), mpow(b, k));
  printf("%d\n", mul_mod(ans, calc(c, (n + 1) / k - 1)));
  return 0;
}

```