#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 9;
const int MAX = 100005;
long long qpow(long long a, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * a) % mod;
    p >>= 1;
    a = (a * a) % mod;
  }
  return ans;
}
long long NY(long long x) { return qpow(x, mod - 2); }
int a, b, k, n;
string s;
int m[MAX];
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long w = qpow(b, k) * NY(qpow(a, k)) % mod;
  w = (w + mod) % mod;
  long long fm = NY(1 - w + mod) % mod;
  long long cnt = (n + 1) / k;
  long long ans = 0;
  for (int i = 0; i < k; ++i) {
    long long res1 = qpow(a, n - i) * qpow(b, i) % mod;
    long long res2 = (1 - qpow(w, cnt) + mod) % mod;
    long long t = ((res1 * res2) % mod * fm) % mod;
    if (w == 1) t = (res1 * cnt) % mod;
    if (s[i] == '-') t *= -1;
    ans = (ans + t) % mod;
  }
  if (ans < 0) ans += mod;
  printf("%I64d\n", ans % mod);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int mod = 1e9 + 9;
const int MAX = 100005;
long long qpow(long long a, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * a) % mod;
    p >>= 1;
    a = (a * a) % mod;
  }
  return ans;
}
long long NY(long long x) { return qpow(x, mod - 2); }
int a, b, k, n;
string s;
int m[MAX];
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  long long w = qpow(b, k) * NY(qpow(a, k)) % mod;
  w = (w + mod) % mod;
  long long fm = NY(1 - w + mod) % mod;
  long long cnt = (n + 1) / k;
  long long ans = 0;
  for (int i = 0; i < k; ++i) {
    long long res1 = qpow(a, n - i) * qpow(b, i) % mod;
    long long res2 = (1 - qpow(w, cnt) + mod) % mod;
    long long t = ((res1 * res2) % mod * fm) % mod;
    if (w == 1) t = (res1 * cnt) % mod;
    if (s[i] == '-') t *= -1;
    ans = (ans + t) % mod;
  }
  if (ans < 0) ans += mod;
  printf("%I64d\n", ans % mod);
  return 0;
}

```