#include <bits/stdc++.h>
long long read() {
  long long res = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    res = res * 10 + ch - 48;
    ch = getchar();
  }
  return res * flag;
}
void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
const long long p = 1e9 + 9.5;
long long a, b, n, k, ans = 0;
std::string s;
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) (res *= a) %= p;
    (a *= a) %= p;
    b >>= 1;
  }
  return res % p;
}
long long mul(long long a, long long b) {
  long long res = 0;
  while (b) {
    if (b & 1) (res += a) %= p;
    (a += a) %= p;
    b >>= 1;
  }
  return res % p;
}
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, y, x);
  y -= (a / b) * x;
}
long long inv(long long a) { return (qpow(a, p - 2) + p) % p; }
void solution() {
  std::cin >> n >> a >> b >> k >> s;
  long long q = mul(qpow(b, k), inv(qpow(a, k)));
  42;
  42;
  42;
  for (long long i = (0); i <= (k - 1); ++i)
    ans = (ans + mul(qpow(a, n - i), qpow(b, i)) * (s[i] == '+' ? 1 : -1) + p) %
          p;
  42;
  print(q == 1 ? ans * ((n + 1) / k) % p
               : mul(inv(1 - q),
                     ((ans - mul(ans, qpow(q, (n + 1) / k))) % p + p) % p)),
      putchar('\n');
}
signed main() { solution(); }

### Prompt

Please create a solution in CPP 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>
long long read() {
  long long res = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    res = res * 10 + ch - 48;
    ch = getchar();
  }
  return res * flag;
}
void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
const long long p = 1e9 + 9.5;
long long a, b, n, k, ans = 0;
std::string s;
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) (res *= a) %= p;
    (a *= a) %= p;
    b >>= 1;
  }
  return res % p;
}
long long mul(long long a, long long b) {
  long long res = 0;
  while (b) {
    if (b & 1) (res += a) %= p;
    (a += a) %= p;
    b >>= 1;
  }
  return res % p;
}
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, y, x);
  y -= (a / b) * x;
}
long long inv(long long a) { return (qpow(a, p - 2) + p) % p; }
void solution() {
  std::cin >> n >> a >> b >> k >> s;
  long long q = mul(qpow(b, k), inv(qpow(a, k)));
  42;
  42;
  42;
  for (long long i = (0); i <= (k - 1); ++i)
    ans = (ans + mul(qpow(a, n - i), qpow(b, i)) * (s[i] == '+' ? 1 : -1) + p) %
          p;
  42;
  print(q == 1 ? ans * ((n + 1) / k) % p
               : mul(inv(1 - q),
                     ((ans - mul(ans, qpow(q, (n + 1) / k))) % p + p) % p)),
      putchar('\n');
}
signed main() { solution(); }

```