#include <bits/stdc++.h>
using namespace std;
long long mod = (long long)1e9 + 9;
long long bigmod(long long n, long long p) {
  if (p == 0) return 1LL;
  long long ans1, ans2 = bigmod(n, p / 2);
  ans1 = (ans2 * ans2) % mod;
  if (p % 2) {
    ans1 = (ans1 * n) % mod;
  }
  return ans1;
}
int main() {
  long long i, j, k, n, m, p, q, a, b;
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  string str;
  cin >> str;
  long long temp, temp1 = bigmod(b, k), temp2 = bigmod(a, k);
  long long r = (temp1 * (bigmod(temp2, mod - 2))) % mod;
  long long ans = 0;
  long long l = str.length();
  for (i = 0; i < l; i++) {
    temp1 = bigmod(a, n - i);
    temp2 = bigmod(b, i);
    temp = (temp1 * temp2) % mod;
    if (str[i] == '+') {
      ans += temp;
      if (ans > mod) ans -= mod;
    } else {
      ans -= temp;
      if (ans < 0) ans += mod;
    }
  }
  temp = (n + 1) / k;
  if (r == 1) {
    ans = (ans * temp) % mod;
    printf("%lld\n", ans);
    return 0;
  }
  long long res;
  temp1 = bigmod(r, temp);
  temp1--;
  if (temp1 < 0) temp1 += mod;
  temp2 = r - 1;
  if (temp2 < 0) temp2 += mod;
  res = (temp1 * (bigmod(temp2, mod - 2))) % mod;
  ans = (ans * res) % mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create 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 mod = (long long)1e9 + 9;
long long bigmod(long long n, long long p) {
  if (p == 0) return 1LL;
  long long ans1, ans2 = bigmod(n, p / 2);
  ans1 = (ans2 * ans2) % mod;
  if (p % 2) {
    ans1 = (ans1 * n) % mod;
  }
  return ans1;
}
int main() {
  long long i, j, k, n, m, p, q, a, b;
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  string str;
  cin >> str;
  long long temp, temp1 = bigmod(b, k), temp2 = bigmod(a, k);
  long long r = (temp1 * (bigmod(temp2, mod - 2))) % mod;
  long long ans = 0;
  long long l = str.length();
  for (i = 0; i < l; i++) {
    temp1 = bigmod(a, n - i);
    temp2 = bigmod(b, i);
    temp = (temp1 * temp2) % mod;
    if (str[i] == '+') {
      ans += temp;
      if (ans > mod) ans -= mod;
    } else {
      ans -= temp;
      if (ans < 0) ans += mod;
    }
  }
  temp = (n + 1) / k;
  if (r == 1) {
    ans = (ans * temp) % mod;
    printf("%lld\n", ans);
    return 0;
  }
  long long res;
  temp1 = bigmod(r, temp);
  temp1--;
  if (temp1 < 0) temp1 += mod;
  temp2 = r - 1;
  if (temp2 < 0) temp2 += mod;
  res = (temp1 * (bigmod(temp2, mod - 2))) % mod;
  ans = (ans * res) % mod;
  printf("%lld\n", ans);
  return 0;
}

```