#include <bits/stdc++.h>
using namespace std;
long long MOD = 1e9 + 9;
long long fpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x % MOD;
  long long p = fpow(x, y / 2) % MOD;
  p *= p;
  p %= MOD;
  if (y % 2) p *= x;
  return p % MOD;
}
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    char c;
    cin >> c;
    if (c == '+')
      ans += fpow(a, n - i) * fpow(b, i);
    else
      ans -= fpow(a, n - i) * fpow(b, i);
    ans %= MOD;
  }
  long long o = (n + 1) / k;
  long long u = b * (fpow(a, MOD - 2));
  u %= MOD;
  u = fpow(u, k);
  u %= MOD;
  if (u == 1) {
    long long sum = ans * o;
    sum %= MOD;
    if (sum < 0) sum += MOD;
    cout << sum << endl;
    return 0;
  }
  long long p = fpow(u, o);
  p %= MOD;
  p -= 1;
  u -= 1;
  u %= MOD;
  long long y = p * fpow(u, MOD - 2);
  y %= MOD;
  long long sum = ans * y;
  sum %= MOD;
  if (sum < 0) sum += MOD;
  cout << sum << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long MOD = 1e9 + 9;
long long fpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x % MOD;
  long long p = fpow(x, y / 2) % MOD;
  p *= p;
  p %= MOD;
  if (y % 2) p *= x;
  return p % MOD;
}
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    char c;
    cin >> c;
    if (c == '+')
      ans += fpow(a, n - i) * fpow(b, i);
    else
      ans -= fpow(a, n - i) * fpow(b, i);
    ans %= MOD;
  }
  long long o = (n + 1) / k;
  long long u = b * (fpow(a, MOD - 2));
  u %= MOD;
  u = fpow(u, k);
  u %= MOD;
  if (u == 1) {
    long long sum = ans * o;
    sum %= MOD;
    if (sum < 0) sum += MOD;
    cout << sum << endl;
    return 0;
  }
  long long p = fpow(u, o);
  p %= MOD;
  p -= 1;
  u -= 1;
  u %= MOD;
  long long y = p * fpow(u, MOD - 2);
  y %= MOD;
  long long sum = ans * y;
  sum %= MOD;
  if (sum < 0) sum += MOD;
  cout << sum << endl;
  return 0;
}

```