#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
long long a, b, n, k;
const long long mod = 1e9 + 9;
char s[maxn];
long long qpow(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * x % mod;
    n >>= 1;
    x = x * x % mod;
  }
  return ans % mod;
}
long long inv(long long x) { return qpow(x, mod - 2); }
void input() {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", s);
}
void slove() {
  long long ans = 0, base = 0;
  for (int i = 0; i < k; ++i) {
    long long tem = qpow(a, n - i) * qpow(b, i);
    if (s[i] == '+')
      base = (base + tem) % mod;
    else
      base = (base + mod - tem) % mod;
  }
  long long tim = (n + 1) / k;
  long long q = qpow(b, k) * inv(qpow(a, k)) % mod;
  if (q == 1)
    ans = tim * base % mod;
  else
    ans = (base * (qpow(q, tim) - 1 + mod) % mod * inv((q - 1 + mod) % mod)) %
          mod;
  printf("%lld\n", (ans + mod) % mod);
}
signed main() {
  input();
  slove();
  return 0;
}

### Prompt

Develop a solution in CPP 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 maxn = 1e5 + 5;
long long a, b, n, k;
const long long mod = 1e9 + 9;
char s[maxn];
long long qpow(long long x, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * x % mod;
    n >>= 1;
    x = x * x % mod;
  }
  return ans % mod;
}
long long inv(long long x) { return qpow(x, mod - 2); }
void input() {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", s);
}
void slove() {
  long long ans = 0, base = 0;
  for (int i = 0; i < k; ++i) {
    long long tem = qpow(a, n - i) * qpow(b, i);
    if (s[i] == '+')
      base = (base + tem) % mod;
    else
      base = (base + mod - tem) % mod;
  }
  long long tim = (n + 1) / k;
  long long q = qpow(b, k) * inv(qpow(a, k)) % mod;
  if (q == 1)
    ans = tim * base % mod;
  else
    ans = (base * (qpow(q, tim) - 1 + mod) % mod * inv((q - 1 + mod) % mod)) %
          mod;
  printf("%lld\n", (ans + mod) % mod);
}
signed main() {
  input();
  slove();
  return 0;
}

```