#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const long long mod = 1e9 + 9;
long long n, a, b, k;
char s[N];
inline long long Pow(long long x, int k) {
  long long y = 1;
  for (; k; k >>= 1, x = x * x % mod)
    if (k & 1) y = y * x % mod;
  return y;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  int tmp = x;
  x = y;
  y = tmp - a / b * y;
}
inline long long inv(int a) {
  long long x, y;
  exgcd(a, mod, x, y);
  return (x % mod + mod) % mod;
}
long long bas;
inline void init() {
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", s);
}
inline void solve() {
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      bas = (bas + 1ll * Pow(a, n - i) * Pow(b, i) % mod + mod) % mod;
    else
      bas = (bas - 1ll * Pow(a, n - i) * Pow(b, i) % mod + mod) % mod;
  }
  long long len = (n + 1) / k, q = Pow(b, k) * Pow(inv(a), k) % mod, ans;
  if (q != 1)
    ans = bas * (Pow(q, len) - 1 + mod) % mod * inv((q - 1 + mod) % mod) % mod;
  else
    ans = bas * len % mod;
  printf("%I64d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}

### Prompt

Your task is to create 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 int N = 1e5 + 5;
const long long mod = 1e9 + 9;
long long n, a, b, k;
char s[N];
inline long long Pow(long long x, int k) {
  long long y = 1;
  for (; k; k >>= 1, x = x * x % mod)
    if (k & 1) y = y * x % mod;
  return y;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  int tmp = x;
  x = y;
  y = tmp - a / b * y;
}
inline long long inv(int a) {
  long long x, y;
  exgcd(a, mod, x, y);
  return (x % mod + mod) % mod;
}
long long bas;
inline void init() {
  scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k);
  scanf("%s", s);
}
inline void solve() {
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      bas = (bas + 1ll * Pow(a, n - i) * Pow(b, i) % mod + mod) % mod;
    else
      bas = (bas - 1ll * Pow(a, n - i) * Pow(b, i) % mod + mod) % mod;
  }
  long long len = (n + 1) / k, q = Pow(b, k) * Pow(inv(a), k) % mod, ans;
  if (q != 1)
    ans = bas * (Pow(q, len) - 1 + mod) % mod * inv((q - 1 + mod) % mod) % mod;
  else
    ans = bas * len % mod;
  printf("%I64d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}

```