#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const int mo = 1000000009;
long long n, a, b, k;
long long s[maxn];
long long sum;
long long qpow(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long ans = qpow(a, b / 2);
  ans = ans * ans;
  ans %= mo;
  if (b % 2 == 1) ans *= a;
  ans %= mo;
  return ans;
}
long long solve(long long x, long long k) {
  if (k == 0) return 1;
  if (k == 1) return (x + 1) % mo;
  long long ans = solve(x, (k - 1) / 2);
  long long tmp = qpow(x, k / 2 + 1);
  ans = ans * (1LL + tmp);
  ans %= mo;
  if (k % 2 == 0) ans += qpow(x, k / 2);
  ans %= mo;
  return ans;
}
int main() {
  cin >> n >> a >> b >> k;
  for (int i = 0; i < k; i++) {
    char ch = getchar();
    while (ch != '+' && ch != '-') ch = getchar();
    s[i] = (ch == '+' ? 1 : -1);
  }
  sum = 0;
  for (int i = 0; i < k; i++) {
    long long tmp = qpow(a, n - i) * qpow(b, i);
    tmp %= mo;
    tmp *= s[i];
    while (tmp < 0) tmp += mo;
    tmp %= mo;
    sum += tmp;
    sum %= mo;
  }
  long long tmp = qpow(qpow(a, k), mo - 2);
  long long x = tmp * qpow(b, k);
  x %= mo;
  long long ans = solve(x, (n + 1) / k - 1);
  ans *= sum;
  ans %= mo;
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 maxn = 100005;
const int mo = 1000000009;
long long n, a, b, k;
long long s[maxn];
long long sum;
long long qpow(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long ans = qpow(a, b / 2);
  ans = ans * ans;
  ans %= mo;
  if (b % 2 == 1) ans *= a;
  ans %= mo;
  return ans;
}
long long solve(long long x, long long k) {
  if (k == 0) return 1;
  if (k == 1) return (x + 1) % mo;
  long long ans = solve(x, (k - 1) / 2);
  long long tmp = qpow(x, k / 2 + 1);
  ans = ans * (1LL + tmp);
  ans %= mo;
  if (k % 2 == 0) ans += qpow(x, k / 2);
  ans %= mo;
  return ans;
}
int main() {
  cin >> n >> a >> b >> k;
  for (int i = 0; i < k; i++) {
    char ch = getchar();
    while (ch != '+' && ch != '-') ch = getchar();
    s[i] = (ch == '+' ? 1 : -1);
  }
  sum = 0;
  for (int i = 0; i < k; i++) {
    long long tmp = qpow(a, n - i) * qpow(b, i);
    tmp %= mo;
    tmp *= s[i];
    while (tmp < 0) tmp += mo;
    tmp %= mo;
    sum += tmp;
    sum %= mo;
  }
  long long tmp = qpow(qpow(a, k), mo - 2);
  long long x = tmp * qpow(b, k);
  x %= mo;
  long long ans = solve(x, (n + 1) / k - 1);
  ans *= sum;
  ans %= mo;
  cout << ans << endl;
  return 0;
}

```