#include <bits/stdc++.h>
using namespace std;
const long long mo = 1e9 + 9;
long long mi(long long a, long long b) {
  long long ans = 1;
  a %= mo;
  while (b) {
    if (b & 1) ans = ans * a % mo;
    b >>= 1;
    a = a * a % mo;
  }
  return ans;
}
char ch[200000];
void ext_gcd(long long a, long long b, long long &d, long long &x,
             long long &y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    ext_gcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long mod_inverse(long long a, long long m) {
  long long x, y, d;
  ext_gcd(a, mo, d, x, y);
  return (mo + x % mo) % mo;
}
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  long long cnt = 1ll * (n + 1) / k;
  scanf("%s", ch);
  long long a1 = 0;
  for (int i = 0; i < k; ++i) {
    if (ch[i] == '-')
      a1 = (a1 - mi(1ll * a, 1ll * (n - i)) * mi(1ll * b, 1ll * i) % mo + mo) %
           mo;
    else
      a1 = (a1 + mi(1ll * a, 1ll * (n - i)) * mi(1ll * b, 1ll * i) % mo + mo) %
           mo;
  }
  long long q = mi(b, k) % mo * mod_inverse(mi(a, k), mo) % mo;
  if (q == 1)
    cout << (a1 * cnt) % mo;
  else {
    long long ans =
        a1 * (1 - mi(q, cnt) + mo) % mo * mod_inverse((1 - q + mo), mo) % mo;
    cout << (ans + mo) % mo;
  }
  return 0;
}

### Prompt

Generate 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;
long long mi(long long a, long long b) {
  long long ans = 1;
  a %= mo;
  while (b) {
    if (b & 1) ans = ans * a % mo;
    b >>= 1;
    a = a * a % mo;
  }
  return ans;
}
char ch[200000];
void ext_gcd(long long a, long long b, long long &d, long long &x,
             long long &y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    ext_gcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long mod_inverse(long long a, long long m) {
  long long x, y, d;
  ext_gcd(a, mo, d, x, y);
  return (mo + x % mo) % mo;
}
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  long long cnt = 1ll * (n + 1) / k;
  scanf("%s", ch);
  long long a1 = 0;
  for (int i = 0; i < k; ++i) {
    if (ch[i] == '-')
      a1 = (a1 - mi(1ll * a, 1ll * (n - i)) * mi(1ll * b, 1ll * i) % mo + mo) %
           mo;
    else
      a1 = (a1 + mi(1ll * a, 1ll * (n - i)) * mi(1ll * b, 1ll * i) % mo + mo) %
           mo;
  }
  long long q = mi(b, k) % mo * mod_inverse(mi(a, k), mo) % mo;
  if (q == 1)
    cout << (a1 * cnt) % mo;
  else {
    long long ans =
        a1 * (1 - mi(q, cnt) + mo) % mo * mod_inverse((1 - q + mo), mo) % mo;
    cout << (ans + mo) % mo;
  }
  return 0;
}

```