#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 9;
const int K = 1e5 + 10;
long long a, b, k, n, inv;
long long aa[K], bb[K];
long long fexp(long long base, long long e) {
  long long ret = 1;
  while (e) {
    if (e & 1) ret = ret * base % mod;
    base = base * base % mod;
    e >>= 1;
  }
  return ret;
}
void cal_ini() {
  aa[0] = fexp(a, n);
  bb[0] = 1;
  inv = fexp(a, mod - 2);
  for (int i = 1; i < k; ++i) {
    aa[i] = aa[i - 1] * inv % mod;
    bb[i] = bb[i - 1] * b % mod;
  }
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  cal_ini();
  long long st = 0;
  long long cr = fexp(inv, k) * fexp(b, k) % mod;
  long long d = (n + 1) / k;
  long long ans;
  getchar();
  for (int i = 0; i < k; ++i) {
    char c = getchar();
    if (c == '+')
      st = (st + aa[i] * bb[i]) % mod;
    else
      st = ((st - aa[i] * bb[i] % mod + mod) % mod);
  }
  if (cr != 1)
    ans = st * fexp(((cr - 1 + mod) % mod), mod - 2) % mod *
          ((fexp(cr, d) - 1 + mod) % mod) % mod;
  else
    ans = st * d % mod;
  printf("%lld\n", ans);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 mod = 1e9 + 9;
const int K = 1e5 + 10;
long long a, b, k, n, inv;
long long aa[K], bb[K];
long long fexp(long long base, long long e) {
  long long ret = 1;
  while (e) {
    if (e & 1) ret = ret * base % mod;
    base = base * base % mod;
    e >>= 1;
  }
  return ret;
}
void cal_ini() {
  aa[0] = fexp(a, n);
  bb[0] = 1;
  inv = fexp(a, mod - 2);
  for (int i = 1; i < k; ++i) {
    aa[i] = aa[i - 1] * inv % mod;
    bb[i] = bb[i - 1] * b % mod;
  }
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  cal_ini();
  long long st = 0;
  long long cr = fexp(inv, k) * fexp(b, k) % mod;
  long long d = (n + 1) / k;
  long long ans;
  getchar();
  for (int i = 0; i < k; ++i) {
    char c = getchar();
    if (c == '+')
      st = (st + aa[i] * bb[i]) % mod;
    else
      st = ((st - aa[i] * bb[i] % mod + mod) % mod);
  }
  if (cr != 1)
    ans = st * fexp(((cr - 1 + mod) % mod), mod - 2) % mod *
          ((fexp(cr, d) - 1 + mod) % mod) % mod;
  else
    ans = st * d % mod;
  printf("%lld\n", ans);
}

```