#include <bits/stdc++.h>
using namespace std;
const long long mod = 1ll * 1e9 + 9ll;
char str[100005];
long long P(long long a, long long b) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1ll) {
      ret = ret * a % mod;
    }
    b /= 2ll;
    a = a * a % mod;
  }
  return ret;
}
int main() {
  long long n, a, b, k;
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", str);
  long long aa = a, bb = b, sum = 0;
  for (int i = 0; i < k; i++) {
    aa = P(a, n - i);
    bb = P(b, i);
    if (str[i] == '+') {
      sum = sum + aa * bb % mod;
      sum %= mod;
    } else {
      sum = sum - aa * bb % mod;
      sum %= mod;
      sum += mod;
      sum %= mod;
    }
  }
  long long tmp = P(P(a, k), mod - 2) * P(b, k) % mod;
  int cnt = (n + 1) / k;
  if (tmp == 1) {
    return !printf("%lld", sum * cnt % mod);
  }
  long long r = tmp;
  r = P(tmp, cnt);
  r = r - 1;
  r %= mod;
  r += mod;
  r %= mod;
  r = r * P((tmp - 1 + mod) % mod, mod - 2) % mod;
  printf("%lld\n", sum * r % mod);
}

### 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;
const long long mod = 1ll * 1e9 + 9ll;
char str[100005];
long long P(long long a, long long b) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1ll) {
      ret = ret * a % mod;
    }
    b /= 2ll;
    a = a * a % mod;
  }
  return ret;
}
int main() {
  long long n, a, b, k;
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", str);
  long long aa = a, bb = b, sum = 0;
  for (int i = 0; i < k; i++) {
    aa = P(a, n - i);
    bb = P(b, i);
    if (str[i] == '+') {
      sum = sum + aa * bb % mod;
      sum %= mod;
    } else {
      sum = sum - aa * bb % mod;
      sum %= mod;
      sum += mod;
      sum %= mod;
    }
  }
  long long tmp = P(P(a, k), mod - 2) * P(b, k) % mod;
  int cnt = (n + 1) / k;
  if (tmp == 1) {
    return !printf("%lld", sum * cnt % mod);
  }
  long long r = tmp;
  r = P(tmp, cnt);
  r = r - 1;
  r %= mod;
  r += mod;
  r %= mod;
  r = r * P((tmp - 1 + mod) % mod, mod - 2) % mod;
  printf("%lld\n", sum * r % mod);
}

```