#include <bits/stdc++.h>
using namespace std;
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
const long long MOD = 1e9 + 9;
long long n, a, b, k, q, r;
char s[100005];
long long pow_(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a % MOD;
  if (b % 2 == 0) return pow_((a * a) % MOD, b / 2) % MOD;
  return (a * pow_((a * a) % MOD, b / 2)) % MOD;
}
long long inv(long long a) { return pow_(a, MOD - 2); }
int main(void) {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", s);
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      s[i] = 1;
    else
      s[i] = -1;
  }
  q = (n + 1) / k;
  long long val = 0;
  long long x = (b * inv(a)) % MOD;
  long long calc = pow_(a, n);
  for (int i = 0; i < k; i++) {
    val = (MOD + val + (long long)s[i] * calc) % MOD;
    calc = (calc * x) % MOD;
  }
  long long tmp_fast = 0;
  long long xk = pow_(x, k);
  if (xk == 1)
    tmp_fast = q;
  else {
    long long val1 = (MOD + pow_(xk, q) - 1) % MOD;
    long long val2 = inv(xk - 1);
    tmp_fast = (val1 * val2) % MOD;
  }
  long long ans = (val * tmp_fast) % MOD;
  printf("%lld", ans);
}

### Prompt

Please formulate 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;
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
const long long MOD = 1e9 + 9;
long long n, a, b, k, q, r;
char s[100005];
long long pow_(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a % MOD;
  if (b % 2 == 0) return pow_((a * a) % MOD, b / 2) % MOD;
  return (a * pow_((a * a) % MOD, b / 2)) % MOD;
}
long long inv(long long a) { return pow_(a, MOD - 2); }
int main(void) {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", s);
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      s[i] = 1;
    else
      s[i] = -1;
  }
  q = (n + 1) / k;
  long long val = 0;
  long long x = (b * inv(a)) % MOD;
  long long calc = pow_(a, n);
  for (int i = 0; i < k; i++) {
    val = (MOD + val + (long long)s[i] * calc) % MOD;
    calc = (calc * x) % MOD;
  }
  long long tmp_fast = 0;
  long long xk = pow_(x, k);
  if (xk == 1)
    tmp_fast = q;
  else {
    long long val1 = (MOD + pow_(xk, q) - 1) % MOD;
    long long val2 = inv(xk - 1);
    tmp_fast = (val1 * val2) % MOD;
  }
  long long ans = (val * tmp_fast) % MOD;
  printf("%lld", ans);
}

```