#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 9;
int n, m, a, b, k;
char s[100005];
long long ans, q;
inline long long f(char a) { return a == '-' ? -1 : 1; }
inline long long fpow(long long a, long long b) {
  long long r = 1;
  for (; b; b >>= 1, a = (a * a) % M)
    if (b & 1) r = (r * a) % M;
  return r;
}
inline long long sum(long long a1, long long q, long long n) {
  if (q == 1) return a1 * n % M;
  long long p = (fpow(q, n) - 1 + M) % M;
  long long qq = (q - 1 + M) % M;
  qq = fpow(qq, M - 2);
  return a1 * p % M * qq % M;
}
int main() {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  q = b * fpow(a, M - 2) % M;
  q = fpow(q, k);
  m = (n + 1) / k;
  scanf("%s", s);
  if (n < k) {
    for (int i = 0; i <= n; i++) {
      long long t = fpow(a, n - i) * fpow(b, i) % M;
      if (s[i] == '-') t = M - t;
      (ans += t) %= M;
    }
    return printf("%lld\n", ans), 0;
  }
  for (int i = 0; i < k; i++) {
    long long x = sum(fpow(a, n - i) * fpow(b, i) % M, q, m);
    if (s[i] == '-') x = M - x;
    (ans += x) %= M;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 M = 1e9 + 9;
int n, m, a, b, k;
char s[100005];
long long ans, q;
inline long long f(char a) { return a == '-' ? -1 : 1; }
inline long long fpow(long long a, long long b) {
  long long r = 1;
  for (; b; b >>= 1, a = (a * a) % M)
    if (b & 1) r = (r * a) % M;
  return r;
}
inline long long sum(long long a1, long long q, long long n) {
  if (q == 1) return a1 * n % M;
  long long p = (fpow(q, n) - 1 + M) % M;
  long long qq = (q - 1 + M) % M;
  qq = fpow(qq, M - 2);
  return a1 * p % M * qq % M;
}
int main() {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  q = b * fpow(a, M - 2) % M;
  q = fpow(q, k);
  m = (n + 1) / k;
  scanf("%s", s);
  if (n < k) {
    for (int i = 0; i <= n; i++) {
      long long t = fpow(a, n - i) * fpow(b, i) % M;
      if (s[i] == '-') t = M - t;
      (ans += t) %= M;
    }
    return printf("%lld\n", ans), 0;
  }
  for (int i = 0; i < k; i++) {
    long long x = sum(fpow(a, n - i) * fpow(b, i) % M, q, m);
    if (s[i] == '-') x = M - x;
    (ans += x) %= M;
  }
  printf("%lld\n", ans);
  return 0;
}

```