#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
void umax(int &a, int b) { a = max(a, b); }
void umin(int &a, int b) { a = min(a, b); }
int dcmp(double x) { return fabs(x) <= 1e-8 ? 0 : (x > 0 ? 1 : -1); }
void file() {
  freopen("data_in.txt", "r", stdin);
  freopen("data_out.txt", "w", stdout);
}
namespace solver {
const long long mod = 1e9 + 9;
long long n, a, b, k;
long long Pow(long long a, long long b) {
  a %= mod;
  b %= mod;
  long long res = 1;
  while (b) {
    if (b & 1) res *= a, res %= mod;
    b >>= 1;
    a *= a;
    a %= mod;
  }
  res += mod;
  res %= mod;
  return res;
}
void solve() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  string str;
  cin >> str;
  long long x = 0;
  for (int i = 0; i < k; i++) {
    if (str[i] == '+') {
      x += (Pow(a, n - i) * Pow(b, i)) % mod;
      x %= mod;
    } else {
      x -= (Pow(a, n - i) * Pow(b, i)) % mod;
      x %= mod;
      x += mod;
      x %= mod;
    }
  }
  n = (n + 1) / k;
  long long q = b * Pow(a, mod - 2) % mod;
  q = Pow(q, k);
  if (q == 1) {
    printf("%lld\n", x * n % mod);
  } else {
    printf("%lld\n",
           ((x * (Pow(q, n) - 1) % mod) % mod * Pow(q - 1, mod - 2)) % mod);
  }
}
}  // namespace solver
int main() {
  solver::solve();
  return 0;
}

### Prompt

In cpp, your task is to solve 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 double pi = acos(-1.0);
void umax(int &a, int b) { a = max(a, b); }
void umin(int &a, int b) { a = min(a, b); }
int dcmp(double x) { return fabs(x) <= 1e-8 ? 0 : (x > 0 ? 1 : -1); }
void file() {
  freopen("data_in.txt", "r", stdin);
  freopen("data_out.txt", "w", stdout);
}
namespace solver {
const long long mod = 1e9 + 9;
long long n, a, b, k;
long long Pow(long long a, long long b) {
  a %= mod;
  b %= mod;
  long long res = 1;
  while (b) {
    if (b & 1) res *= a, res %= mod;
    b >>= 1;
    a *= a;
    a %= mod;
  }
  res += mod;
  res %= mod;
  return res;
}
void solve() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  string str;
  cin >> str;
  long long x = 0;
  for (int i = 0; i < k; i++) {
    if (str[i] == '+') {
      x += (Pow(a, n - i) * Pow(b, i)) % mod;
      x %= mod;
    } else {
      x -= (Pow(a, n - i) * Pow(b, i)) % mod;
      x %= mod;
      x += mod;
      x %= mod;
    }
  }
  n = (n + 1) / k;
  long long q = b * Pow(a, mod - 2) % mod;
  q = Pow(q, k);
  if (q == 1) {
    printf("%lld\n", x * n % mod);
  } else {
    printf("%lld\n",
           ((x * (Pow(q, n) - 1) % mod) % mod * Pow(q - 1, mod - 2)) % mod);
  }
}
}  // namespace solver
int main() {
  solver::solve();
  return 0;
}

```