#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
const int mod = 1000000009;
int n, a, b, k, q, m, a1, ans;
int c[N];
char s[N];
int quickpow(int x, int y) {
  int s = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) s = 1ll * s * x % mod;
  return s;
}
int main() {
  int i;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  q = 1ll * quickpow(a, mod - 2) * b % mod;
  q = quickpow(q, k);
  scanf("%s", s);
  for (i = 0; i < k; i++) {
    if (s[i] == '+')
      c[i] = 1;
    else
      c[i] = -1;
  }
  for (i = 0; i < k; i++) {
    a1 = 1ll * quickpow(a, n - i) * quickpow(b, i) % mod;
    m = n / k + 1;
    if (q == 1) {
      (ans += 1ll * c[i] * a1 % mod * m % mod) %= mod;
      continue;
    }
    (ans += 1ll * c[i] * a1 * (1 - quickpow(q, m)) % mod *
            quickpow(1 - q, mod - 2) % mod) %= mod;
  }
  ans = (ans % mod + mod) % mod;
  printf("%d\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;
const int N = 100005;
const int mod = 1000000009;
int n, a, b, k, q, m, a1, ans;
int c[N];
char s[N];
int quickpow(int x, int y) {
  int s = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod)
    if (y & 1) s = 1ll * s * x % mod;
  return s;
}
int main() {
  int i;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  q = 1ll * quickpow(a, mod - 2) * b % mod;
  q = quickpow(q, k);
  scanf("%s", s);
  for (i = 0; i < k; i++) {
    if (s[i] == '+')
      c[i] = 1;
    else
      c[i] = -1;
  }
  for (i = 0; i < k; i++) {
    a1 = 1ll * quickpow(a, n - i) * quickpow(b, i) % mod;
    m = n / k + 1;
    if (q == 1) {
      (ans += 1ll * c[i] * a1 % mod * m % mod) %= mod;
      continue;
    }
    (ans += 1ll * c[i] * a1 * (1 - quickpow(q, m)) % mod *
            quickpow(1 - q, mod - 2) % mod) %= mod;
  }
  ans = (ans % mod + mod) % mod;
  printf("%d\n", ans);
  return 0;
}

```