#include <bits/stdc++.h>
using namespace std;
const long long infl = 1e18 + 5;
long long n, m, k, q, x, y, f, val, t, i, j;
long long ind, cnt, sz, sm, ans, mx, mn;
long long a, b;
long long modpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res *= a;
      res %= 1000000009;
    }
    a *= a;
    b /= 2;
    a %= 1000000009;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("inp.txt", "r")) {
    freopen("myfile.txt", "w", stdout);
    freopen("inp.txt", "r", stdin);
  }
  string s;
  cin >> n >> a >> b >> k;
  cin >> s;
  val = 0;
  for (i = 0; i < k; i++) {
    x = modpow(a, n - i) * modpow(b, i) % 1000000009;
    if (s[i] == '+')
      val += x;
    else
      val -= x;
    val += 2LL * 1000000009;
    val %= 1000000009;
  }
  y = (n + 1) / k;
  long long r =
      modpow(b, k) * modpow(modpow(a, k), 1000000009 - 2LL) % 1000000009;
  if (r == 1) {
    ans = val * y % 1000000009;
    cout << ans;
    return 0;
  }
  long long num = (modpow(r, y) - 1LL + 1000000009) % 1000000009;
  long long den = (r - 1LL + 1000000009) % 1000000009;
  long long mden = modpow(den, 1000000009 - 2LL);
  val = val * num % 1000000009;
  ans = val * mden % 1000000009;
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 long long infl = 1e18 + 5;
long long n, m, k, q, x, y, f, val, t, i, j;
long long ind, cnt, sz, sm, ans, mx, mn;
long long a, b;
long long modpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res *= a;
      res %= 1000000009;
    }
    a *= a;
    b /= 2;
    a %= 1000000009;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("inp.txt", "r")) {
    freopen("myfile.txt", "w", stdout);
    freopen("inp.txt", "r", stdin);
  }
  string s;
  cin >> n >> a >> b >> k;
  cin >> s;
  val = 0;
  for (i = 0; i < k; i++) {
    x = modpow(a, n - i) * modpow(b, i) % 1000000009;
    if (s[i] == '+')
      val += x;
    else
      val -= x;
    val += 2LL * 1000000009;
    val %= 1000000009;
  }
  y = (n + 1) / k;
  long long r =
      modpow(b, k) * modpow(modpow(a, k), 1000000009 - 2LL) % 1000000009;
  if (r == 1) {
    ans = val * y % 1000000009;
    cout << ans;
    return 0;
  }
  long long num = (modpow(r, y) - 1LL + 1000000009) % 1000000009;
  long long den = (r - 1LL + 1000000009) % 1000000009;
  long long mden = modpow(den, 1000000009 - 2LL);
  val = val * num % 1000000009;
  ans = val * mden % 1000000009;
  cout << ans;
  return 0;
}

```