#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 9;
int n, a, b, k;
long long qp(long long a, long long b, int mod) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
long long inv(long long a) { return qp(a, mod - 2, mod); }
char s[100000 + 8];
int main() {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", s);
  long long curb = 1, cura = 1;
  long long circ = 0;
  long long ans = 0;
  long long rem = 0;
  for (int i = 0; i < min(n + 1, k); ++i) {
    if (n >= k && i == n + 1 - (n + 1) / k * k) rem = circ;
    if (s[i] == '+')
      circ = (circ + (inv(cura) * curb % mod)) % mod;
    else
      circ = (circ - (inv(cura) * curb % mod) + mod) % mod;
    cura = cura * a % mod;
    curb = curb * b % mod;
  }
  if (n + 1 <= k) {
    printf("%I64d\n", circ * qp(a, n, mod) % mod);
    return 0;
  }
  long long bkak = qp(b, k, mod) * inv(qp(a, k, mod)) % mod;
  long long circnt = (n + 1) / k;
  if (bkak > 1)
    ans = (ans + ((((qp(bkak, circnt, mod) - 1 + mod) % mod) *
                   inv((bkak - 1 + mod) % mod) % mod) *
                  circ % mod)) %
          mod;
  else
    ans = (ans + (circnt * circ % mod)) % mod;
  ans = (ans + (rem * qp(bkak, circnt, mod) % mod)) % mod;
  ans = ans * qp(a, n, mod) % mod;
  printf("%I64d\n", ans);
}

### Prompt

Your challenge is to write 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>
using namespace std;
const int mod = 1e9 + 9;
int n, a, b, k;
long long qp(long long a, long long b, int mod) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
long long inv(long long a) { return qp(a, mod - 2, mod); }
char s[100000 + 8];
int main() {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", s);
  long long curb = 1, cura = 1;
  long long circ = 0;
  long long ans = 0;
  long long rem = 0;
  for (int i = 0; i < min(n + 1, k); ++i) {
    if (n >= k && i == n + 1 - (n + 1) / k * k) rem = circ;
    if (s[i] == '+')
      circ = (circ + (inv(cura) * curb % mod)) % mod;
    else
      circ = (circ - (inv(cura) * curb % mod) + mod) % mod;
    cura = cura * a % mod;
    curb = curb * b % mod;
  }
  if (n + 1 <= k) {
    printf("%I64d\n", circ * qp(a, n, mod) % mod);
    return 0;
  }
  long long bkak = qp(b, k, mod) * inv(qp(a, k, mod)) % mod;
  long long circnt = (n + 1) / k;
  if (bkak > 1)
    ans = (ans + ((((qp(bkak, circnt, mod) - 1 + mod) % mod) *
                   inv((bkak - 1 + mod) % mod) % mod) *
                  circ % mod)) %
          mod;
  else
    ans = (ans + (circnt * circ % mod)) % mod;
  ans = (ans + (rem * qp(bkak, circnt, mod) % mod)) % mod;
  ans = ans * qp(a, n, mod) % mod;
  printf("%I64d\n", ans);
}

```