#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k, inv;
long long aa[200005], bb[200005];
const long long mod = 1e9 + 9;
long long powmod(long long n, long long k) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * n % mod;
    n = n * n % mod;
    k >>= 1;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  while (cin >> n >> a >> b >> k) {
    aa[0] = powmod(a, n);
    bb[0] = 1;
    inv = powmod(a, mod - 2);
    for (int i = 1; i < k; i++) {
      aa[i] = aa[i - 1] * inv % mod;
      bb[i] = bb[i - 1] * b % mod;
    }
    long long ans = 0;
    for (int i = 0; i < k; i++) {
      char c;
      cin >> c;
      if (c == '+') {
        ans = (ans + aa[i] * bb[i]) % mod;
      } else {
        ans = (ans - (aa[i] * bb[i] % mod) + mod) % mod;
      }
    }
    long long mul = powmod(inv, k) * powmod(b, k) % mod;
    long long t = (n + 1) / k;
    long long sum;
    if (mul == 1) {
      sum = ans * t % mod;
    } else {
      sum = ans * powmod((mul - 1 + mod) % mod, mod - 2) % mod *
            ((powmod(mul, t) - 1 + mod) % mod) % mod;
    }
    cout << sum << endl;
  }
}

### Prompt

Please formulate 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;
long long n, a, b, k, inv;
long long aa[200005], bb[200005];
const long long mod = 1e9 + 9;
long long powmod(long long n, long long k) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * n % mod;
    n = n * n % mod;
    k >>= 1;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  while (cin >> n >> a >> b >> k) {
    aa[0] = powmod(a, n);
    bb[0] = 1;
    inv = powmod(a, mod - 2);
    for (int i = 1; i < k; i++) {
      aa[i] = aa[i - 1] * inv % mod;
      bb[i] = bb[i - 1] * b % mod;
    }
    long long ans = 0;
    for (int i = 0; i < k; i++) {
      char c;
      cin >> c;
      if (c == '+') {
        ans = (ans + aa[i] * bb[i]) % mod;
      } else {
        ans = (ans - (aa[i] * bb[i] % mod) + mod) % mod;
      }
    }
    long long mul = powmod(inv, k) * powmod(b, k) % mod;
    long long t = (n + 1) / k;
    long long sum;
    if (mul == 1) {
      sum = ans * t % mod;
    } else {
      sum = ans * powmod((mul - 1 + mod) % mod, mod - 2) % mod *
            ((powmod(mul, t) - 1 + mod) % mod) % mod;
    }
    cout << sum << endl;
  }
}

```