#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int MAXN = 1e5 + 5;
const long long MOD = 1e9 + 9;
char s[MAXN];
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ans % MOD;
}
long long inv(long long a) { return qpow(a, MOD - 2) % MOD; }
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  scanf("%s", &s);
  long long xh = (n + 1) / k;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      ans += qpow(a, n - i) * qpow(b, i) % MOD;
      ans %= MOD;
    } else {
      ans -= qpow(a, n - i) * qpow(b, i) % MOD;
      while (ans < 0) ans += MOD;
      ans %= MOD;
    }
  }
  long long q = qpow(b, k) * inv(qpow(a, k)) % MOD;
  if (q == 1) {
    ans *= xh;
    ans %= MOD;
  } else {
    long long fz = ans - ans * qpow(q, xh) % MOD;
    while (fz < 0) fz += MOD;
    fz %= MOD;
    long long fm = 1 - q;
    while (fm < 0) fm += MOD;
    fm %= MOD;
    ans = fz * inv(fm) % MOD;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 double eps = 1e-8;
const int MAXN = 1e5 + 5;
const long long MOD = 1e9 + 9;
char s[MAXN];
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ans % MOD;
}
long long inv(long long a) { return qpow(a, MOD - 2) % MOD; }
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  scanf("%s", &s);
  long long xh = (n + 1) / k;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      ans += qpow(a, n - i) * qpow(b, i) % MOD;
      ans %= MOD;
    } else {
      ans -= qpow(a, n - i) * qpow(b, i) % MOD;
      while (ans < 0) ans += MOD;
      ans %= MOD;
    }
  }
  long long q = qpow(b, k) * inv(qpow(a, k)) % MOD;
  if (q == 1) {
    ans *= xh;
    ans %= MOD;
  } else {
    long long fz = ans - ans * qpow(q, xh) % MOD;
    while (fz < 0) fz += MOD;
    fz %= MOD;
    long long fm = 1 - q;
    while (fm < 0) fm += MOD;
    fm %= MOD;
    ans = fz * inv(fm) % MOD;
  }
  cout << ans << endl;
  return 0;
}

```