#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 9;
long long pow(long long x, long long n) {
  long long ret = 1;
  x = (x + MOD) % MOD;
  while (n) {
    if (n & 1) ret = ret * x % MOD;
    n >>= 1;
    x = x * x % MOD;
  }
  return ret;
}
long long fun(long long a, long long b, long long n) {
  if (n <= 1) return b;
  if (n & 1)
    return (fun(a, b, n - 1) + (b * pow(a, n - 1) % MOD)) % MOD;
  else
    return fun(a * a % MOD, (a + 1) * b % MOD, n >> 1);
}
long long a, b, aa, bb, na, q, ans, ak, s0;
char s[100100];
int main() {
  int n, k;
  scanf("%d%I64d%I64d%d", &n, &a, &b, &k);
  scanf(" %s", s);
  aa = pow(a, n);
  na = pow(a, MOD - 2);
  bb = 1;
  s0 = 0;
  for (long long i = 0; i < k; i++) {
    ak = pow(a, n - i) * pow(b, i) % MOD;
    if (s[i] == '-')
      s0 = (s0 - ak + MOD) % MOD;
    else
      s0 = (s0 + ak) % MOD;
  }
  n = (n + 1) / k;
  q = pow(na, (long long)k) * pow(b, (long long)k) % MOD;
  if (q == 1) {
    ans = s0 * n % MOD;
    printf("%I64d\n", ans);
    return 0;
  }
  ans = pow((q - 1), MOD - 2) * (pow(q, (long long)n) - 1) % MOD * s0 % MOD;
  printf("%I64d\n", ans);
}

### Prompt

Develop a solution in cpp to the problem described below:

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 long long MOD = 1e9 + 9;
long long pow(long long x, long long n) {
  long long ret = 1;
  x = (x + MOD) % MOD;
  while (n) {
    if (n & 1) ret = ret * x % MOD;
    n >>= 1;
    x = x * x % MOD;
  }
  return ret;
}
long long fun(long long a, long long b, long long n) {
  if (n <= 1) return b;
  if (n & 1)
    return (fun(a, b, n - 1) + (b * pow(a, n - 1) % MOD)) % MOD;
  else
    return fun(a * a % MOD, (a + 1) * b % MOD, n >> 1);
}
long long a, b, aa, bb, na, q, ans, ak, s0;
char s[100100];
int main() {
  int n, k;
  scanf("%d%I64d%I64d%d", &n, &a, &b, &k);
  scanf(" %s", s);
  aa = pow(a, n);
  na = pow(a, MOD - 2);
  bb = 1;
  s0 = 0;
  for (long long i = 0; i < k; i++) {
    ak = pow(a, n - i) * pow(b, i) % MOD;
    if (s[i] == '-')
      s0 = (s0 - ak + MOD) % MOD;
    else
      s0 = (s0 + ak) % MOD;
  }
  n = (n + 1) / k;
  q = pow(na, (long long)k) * pow(b, (long long)k) % MOD;
  if (q == 1) {
    ans = s0 * n % MOD;
    printf("%I64d\n", ans);
    return 0;
  }
  ans = pow((q - 1), MOD - 2) * (pow(q, (long long)n) - 1) % MOD * s0 % MOD;
  printf("%I64d\n", ans);
}

```