#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long mod = 1e9 + 9;
const long long INF = 1e18 + 1LL;
const int inf = 1e9 + 1e8;
const double PI = acos(-1.0);
const int N = 2e5 + 100;
long long n, a, b, k;
char s[N];
long long PowMod(long long a, long long b) {
  long long ret = 1;
  a = a % mod;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long q = PowMod(b * PowMod(a, mod - 2), k);
  long long ans = 0;
  for (int i = 0; i < k; ++i) {
    int p = 1;
    if (s[i] == '-') p = -1;
    ans = (ans + p * PowMod(a, n - i) * PowMod(b, i)) % mod;
  }
  if (q == 1) {
    ans = ans * ((n + 1) / k);
    return 0 * printf("%lld\n", ((ans % mod) + mod) % mod);
  }
  ans =
      (((ans * (PowMod(q, (n + 1) / k) - 1)) % mod) * PowMod(q - 1, mod - 2)) %
      mod;
  printf("%lld\n", (ans + mod) % mod);
  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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const long long mod = 1e9 + 9;
const long long INF = 1e18 + 1LL;
const int inf = 1e9 + 1e8;
const double PI = acos(-1.0);
const int N = 2e5 + 100;
long long n, a, b, k;
char s[N];
long long PowMod(long long a, long long b) {
  long long ret = 1;
  a = a % mod;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ret;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long q = PowMod(b * PowMod(a, mod - 2), k);
  long long ans = 0;
  for (int i = 0; i < k; ++i) {
    int p = 1;
    if (s[i] == '-') p = -1;
    ans = (ans + p * PowMod(a, n - i) * PowMod(b, i)) % mod;
  }
  if (q == 1) {
    ans = ans * ((n + 1) / k);
    return 0 * printf("%lld\n", ((ans % mod) + mod) % mod);
  }
  ans =
      (((ans * (PowMod(q, (n + 1) / k) - 1)) % mod) * PowMod(q - 1, mod - 2)) %
      mod;
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}

```