#include <bits/stdc++.h>
using namespace std;
template <typename _T>
inline void read(_T &f) {
  f = 0;
  _T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
const long long P = 1e9 + 9;
const int N = 1e5 + 5;
long long power(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % P;
    y >>= 1;
    x = x * x % P;
  }
  return ans;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - a / b * y;
  return d;
}
long long inv(long long a) {
  long long x, y;
  exgcd(a, P, x, y);
  return (x % P + P) % P;
}
char c[N];
long long n, a, b, k, x, y;
int main() {
  cin >> n >> a >> b >> k;
  swap(a, b);
  y = a * inv(b) % P;
  scanf("%s", c + 1);
  long long now = power(b, n);
  x = now * (c[1] == '+' ? 1ll : -1ll);
  for (int i = 2; i <= k; i++) {
    now = now * y % P;
    long long fu = (c[i] == '+' ? 1ll : -1ll);
    x = ((x + now * fu) % P + P) % P;
  }
  y = power(y, k);
  if (power(a, k) == power(b, k)) {
    cout << ((x * (n + 1ll) / k) % P + P) % P << endl;
    return 0;
  }
  cout << ((x * (power(y, n / k + 1ll) - 1ll) % P * inv(y - 1ll) % P + P) % P)
       << endl;
  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;
template <typename _T>
inline void read(_T &f) {
  f = 0;
  _T fu = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fu = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    f = (f << 3) + (f << 1) + (c & 15);
    c = getchar();
  }
  f *= fu;
}
const long long P = 1e9 + 9;
const int N = 1e5 + 5;
long long power(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % P;
    y >>= 1;
    x = x * x % P;
  }
  return ans;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - a / b * y;
  return d;
}
long long inv(long long a) {
  long long x, y;
  exgcd(a, P, x, y);
  return (x % P + P) % P;
}
char c[N];
long long n, a, b, k, x, y;
int main() {
  cin >> n >> a >> b >> k;
  swap(a, b);
  y = a * inv(b) % P;
  scanf("%s", c + 1);
  long long now = power(b, n);
  x = now * (c[1] == '+' ? 1ll : -1ll);
  for (int i = 2; i <= k; i++) {
    now = now * y % P;
    long long fu = (c[i] == '+' ? 1ll : -1ll);
    x = ((x + now * fu) % P + P) % P;
  }
  y = power(y, k);
  if (power(a, k) == power(b, k)) {
    cout << ((x * (n + 1ll) / k) % P + P) % P << endl;
    return 0;
  }
  cout << ((x * (power(y, n / k + 1ll) - 1ll) % P * inv(y - 1ll) % P + P) % P)
       << endl;
  return 0;
}

```