#include <bits/stdc++.h>
using namespace std;
const long long mo = 1e9 + 9;
const int maxn = 1e5 + 10;
long long n, a, b, k;
char ch[maxn];
long long find(long long a, long long t) {
  long long res = 1;
  while (t) {
    if (t & 1) res = res * a % mo;
    a = (a * a) % mo;
    t /= 2;
  }
  return (res + mo) % mo;
}
long long ex_gcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long te = ex_gcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - (a / b) * x;
  return te;
}
long long NY(long long nu) {
  long long x, y;
  ex_gcd(nu, mo, x, y);
  return (x + mo) % mo;
}
int main() {
  int i, j;
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", ch);
  long long ans = 0, m = (n + 1) / k;
  for (i = 0; i < k; i++) {
    long long t1 = find(a, n - i), t2 = find(b, i);
    long long temp = t1 * t2 % mo;
    if (ch[i] == '+')
      ans += temp;
    else
      ans -= temp;
    ans += mo;
    ans %= mo;
  }
  long long q = find(NY(a), k) * find(b, k) % mo;
  if (q == 1) {
    ans = ans * m % mo;
  } else {
    long long t1 = NY(q - 1) % mo;
    ans = (ans * (find(q, m) - 1) % mo) * t1 % mo;
    ans %= mo;
  }
  ans = (ans + mo) % mo;
  printf("%I64d\n", ans);
  return 0;
}

### 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 long long mo = 1e9 + 9;
const int maxn = 1e5 + 10;
long long n, a, b, k;
char ch[maxn];
long long find(long long a, long long t) {
  long long res = 1;
  while (t) {
    if (t & 1) res = res * a % mo;
    a = (a * a) % mo;
    t /= 2;
  }
  return (res + mo) % mo;
}
long long ex_gcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long te = ex_gcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - (a / b) * x;
  return te;
}
long long NY(long long nu) {
  long long x, y;
  ex_gcd(nu, mo, x, y);
  return (x + mo) % mo;
}
int main() {
  int i, j;
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", ch);
  long long ans = 0, m = (n + 1) / k;
  for (i = 0; i < k; i++) {
    long long t1 = find(a, n - i), t2 = find(b, i);
    long long temp = t1 * t2 % mo;
    if (ch[i] == '+')
      ans += temp;
    else
      ans -= temp;
    ans += mo;
    ans %= mo;
  }
  long long q = find(NY(a), k) * find(b, k) % mo;
  if (q == 1) {
    ans = ans * m % mo;
  } else {
    long long t1 = NY(q - 1) % mo;
    ans = (ans * (find(q, m) - 1) % mo) * t1 % mo;
    ans %= mo;
  }
  ans = (ans + mo) % mo;
  printf("%I64d\n", ans);
  return 0;
}

```