#include <bits/stdc++.h>
using namespace std;
long long int arr[100100];
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(long long int a, long long int m) {
  long long int x, y, res;
  a = ((a % 1000000009) + 1000000009) % m;
  long long int g = gcdExtended(a, m, &x, &y);
  if (g != 1) {
    cout << a << " = a " << m << " = m Inverse doesn't exist\n";
    return 0;
  } else {
    res = (x % m + m) % m;
  }
  return res;
}
long long int power(long long int x, long long int n) {
  long long int result = 1;
  while (n) {
    if (n & 1) result = result * x % 1000000009;
    n = n / 2;
    x = x * x % 1000000009;
  }
  result = (result % 1000000009 + 1000000009) % 1000000009;
  return result;
}
long long int sol(long long int a, long long int r1, long long int r2,
                  long long int n) {
  if (r1 != r2) {
    return (((a % 1000000009 *
              (power(r1, n) % 1000000009 - power(r2, n) % 1000000009) %
              1000000009) %
             1000000009 *
             modInverse((power(r2, n - 1) % 1000000009 *
                         (r1 % 1000000009 - r2 % 1000000009) % 1000000009) %
                            1000000009,
                        1000000009)) %
                1000000009 +
            1000000009) %
           1000000009;
  } else
    return ((((a % 1000000009) * (n % 1000000009)) % 1000000009) % 1000000009 +
            1000000009) %
           1000000009;
}
int main() {
  long long int n, a, b, k, ans = 0;
  char ch;
  cin >> n >> a >> b >> k;
  for (long long int i = 0; i < k; i++) {
    cin >> ch;
    if (ch == '+')
      arr[i] = 1;
    else if (ch == '-')
      arr[i] = -1;
  }
  for (long long int i = 0; i < k; i++) {
    ans = (ans % 1000000009 +
           arr[i] *
               sol((power(a, n - i) % 1000000009 * power(b, i) % 1000000009) %
                       1000000009,
                   power(b, k), power(a, k), (n + 1) / k) %
               1000000009) %
          1000000009;
    ans = (ans % 1000000009 + 1000000009) % 1000000009;
  }
  cout << 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;
long long int arr[100100];
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(long long int a, long long int m) {
  long long int x, y, res;
  a = ((a % 1000000009) + 1000000009) % m;
  long long int g = gcdExtended(a, m, &x, &y);
  if (g != 1) {
    cout << a << " = a " << m << " = m Inverse doesn't exist\n";
    return 0;
  } else {
    res = (x % m + m) % m;
  }
  return res;
}
long long int power(long long int x, long long int n) {
  long long int result = 1;
  while (n) {
    if (n & 1) result = result * x % 1000000009;
    n = n / 2;
    x = x * x % 1000000009;
  }
  result = (result % 1000000009 + 1000000009) % 1000000009;
  return result;
}
long long int sol(long long int a, long long int r1, long long int r2,
                  long long int n) {
  if (r1 != r2) {
    return (((a % 1000000009 *
              (power(r1, n) % 1000000009 - power(r2, n) % 1000000009) %
              1000000009) %
             1000000009 *
             modInverse((power(r2, n - 1) % 1000000009 *
                         (r1 % 1000000009 - r2 % 1000000009) % 1000000009) %
                            1000000009,
                        1000000009)) %
                1000000009 +
            1000000009) %
           1000000009;
  } else
    return ((((a % 1000000009) * (n % 1000000009)) % 1000000009) % 1000000009 +
            1000000009) %
           1000000009;
}
int main() {
  long long int n, a, b, k, ans = 0;
  char ch;
  cin >> n >> a >> b >> k;
  for (long long int i = 0; i < k; i++) {
    cin >> ch;
    if (ch == '+')
      arr[i] = 1;
    else if (ch == '-')
      arr[i] = -1;
  }
  for (long long int i = 0; i < k; i++) {
    ans = (ans % 1000000009 +
           arr[i] *
               sol((power(a, n - i) % 1000000009 * power(b, i) % 1000000009) %
                       1000000009,
                   power(b, k), power(a, k), (n + 1) / k) %
               1000000009) %
          1000000009;
    ans = (ans % 1000000009 + 1000000009) % 1000000009;
  }
  cout << ans;
}

```