#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 9;
long long modexp(long long x, long long n) {
  if (n == 0) return 1;
  if (n == 1) return x;
  long long ret = modexp(x, n / 2);
  ret *= ret;
  ret %= mod;
  if (n % 2 == 1) {
    ret *= x;
    ret %= mod;
  }
  if (ret < 0) ret += mod;
  return ret;
}
long long ans;
long long n, k, a, b;
string s;
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  for (long long i = 0; i < k; i++) {
    long long ret = modexp(a, n - i) * modexp(b, i);
    ret %= mod;
    if (s[i] == '+')
      ans += ret;
    else
      ans -= ret;
    if (ans < 0) ans += mod;
    ans %= mod;
  }
  long long r = (n + 1) / k;
  if (a == b) {
    ans *= r;
    ans %= mod;
    cout << ans;
    return 0;
  }
  long long ba = (b * modexp(a, mod - 2)) % mod;
  long long p = modexp(ba, k);
  long long up;
  if (p % mod == 1) {
    up = r;
  } else {
    long long pr = modexp(p, r) - 1;
    if (pr < 0) pr += mod;
    long long dp = modexp((p - 1 + mod) % mod, mod - 2);
    up = (pr * dp) % mod;
    if (up < 0) up += mod;
  }
  ans *= up;
  ans = (ans % mod + mod) % mod;
  cout << ans;
  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;
long long mod = 1e9 + 9;
long long modexp(long long x, long long n) {
  if (n == 0) return 1;
  if (n == 1) return x;
  long long ret = modexp(x, n / 2);
  ret *= ret;
  ret %= mod;
  if (n % 2 == 1) {
    ret *= x;
    ret %= mod;
  }
  if (ret < 0) ret += mod;
  return ret;
}
long long ans;
long long n, k, a, b;
string s;
int main() {
  cin >> n >> a >> b >> k;
  cin >> s;
  for (long long i = 0; i < k; i++) {
    long long ret = modexp(a, n - i) * modexp(b, i);
    ret %= mod;
    if (s[i] == '+')
      ans += ret;
    else
      ans -= ret;
    if (ans < 0) ans += mod;
    ans %= mod;
  }
  long long r = (n + 1) / k;
  if (a == b) {
    ans *= r;
    ans %= mod;
    cout << ans;
    return 0;
  }
  long long ba = (b * modexp(a, mod - 2)) % mod;
  long long p = modexp(ba, k);
  long long up;
  if (p % mod == 1) {
    up = r;
  } else {
    long long pr = modexp(p, r) - 1;
    if (pr < 0) pr += mod;
    long long dp = modexp((p - 1 + mod) % mod, mod - 2);
    up = (pr * dp) % mod;
    if (up < 0) up += mod;
  }
  ans *= up;
  ans = (ans % mod + mod) % mod;
  cout << ans;
  return 0;
}

```