#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k, x, m, A, B, ans;
char s[100010];
string str;
long long expo(long long a, long long b) {
  a %= 1000000009;
  if (!b) return 1;
  if (b % 2) return (a * expo(a * a, b / 2)) % 1000000009;
  return expo(a * a, b / 2);
}
long long invmod(long long x) { return expo(x, 1000000009 - 2); }
long long mul(long long a, long long b) {
  a %= 1000000009;
  b %= 1000000009;
  return (a * b) % 1000000009;
}
int main() {
  scanf("%lld %lld %lld %lld %s", &n, &A, &B, &k, s);
  str = s;
  a = expo(A, n);
  b = 1;
  m = mul(expo(B, k), invmod(expo(A, k)));
  for (int i = 0; i < k; i++) {
    long long val =
        (mul(a, b) * (str[i] == '+' ? 1 : -1) + 1000000009) % 1000000009;
    x = (x + val) % 1000000009;
    a = (a * invmod(A)) % 1000000009;
    b = (b * B) % 1000000009;
  }
  if (m == 1) {
    ans = mul(x, (n + 1) / k);
  } else {
    ans = (1 - expo(m, (n + 1) / k) + 1000000009) % 1000000009;
    ans = (ans * invmod((1 - m + 1000000009) % 1000000009)) % 1000000009;
    ans = mul(ans, x);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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, x, m, A, B, ans;
char s[100010];
string str;
long long expo(long long a, long long b) {
  a %= 1000000009;
  if (!b) return 1;
  if (b % 2) return (a * expo(a * a, b / 2)) % 1000000009;
  return expo(a * a, b / 2);
}
long long invmod(long long x) { return expo(x, 1000000009 - 2); }
long long mul(long long a, long long b) {
  a %= 1000000009;
  b %= 1000000009;
  return (a * b) % 1000000009;
}
int main() {
  scanf("%lld %lld %lld %lld %s", &n, &A, &B, &k, s);
  str = s;
  a = expo(A, n);
  b = 1;
  m = mul(expo(B, k), invmod(expo(A, k)));
  for (int i = 0; i < k; i++) {
    long long val =
        (mul(a, b) * (str[i] == '+' ? 1 : -1) + 1000000009) % 1000000009;
    x = (x + val) % 1000000009;
    a = (a * invmod(A)) % 1000000009;
    b = (b * B) % 1000000009;
  }
  if (m == 1) {
    ans = mul(x, (n + 1) / k);
  } else {
    ans = (1 - expo(m, (n + 1) / k) + 1000000009) % 1000000009;
    ans = (ans * invmod((1 - m + 1000000009) % 1000000009)) % 1000000009;
    ans = mul(ans, x);
  }
  printf("%lld\n", ans);
  return 0;
}

```