#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 9;
int n, a, b, k;
int s[100005];
long long power(long long x, long long t) {
  if (t == 0) return 1LL;
  long long cc = power(x, t >> 1);
  cc = cc * cc % MOD;
  if (t & 1) cc = cc * x % MOD;
  return cc;
}
long long divi(long long a, long long b) { return a * power(b, MOD - 2) % MOD; }
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  getchar();
  for (int i = 0; i < (int)(k); i++) s[i] = getchar() == '+' ? 1 : -1;
  long long cur = 0LL;
  for (int i = 0; i < (int)(k); i++)
    cur = (cur + power(a, k - 1 - i) * power(b, i) * s[i]) % MOD;
  cur = (cur + MOD) % MOD;
  long long ca = power(a, k), cb = power(b, k);
  int ct = (n + 1) / k;
  if (ca == cb)
    cur *= power(cb, ct - 1) * ct;
  else
    cur *= power(cb, ct - 1) *
           divi(1 - power(divi(ca, cb), ct), 1 - divi(ca, cb)) % MOD;
  cur = (cur % MOD + MOD) % MOD;
  printf("%d\n", int(cur));
  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 long long MOD = 1e9 + 9;
int n, a, b, k;
int s[100005];
long long power(long long x, long long t) {
  if (t == 0) return 1LL;
  long long cc = power(x, t >> 1);
  cc = cc * cc % MOD;
  if (t & 1) cc = cc * x % MOD;
  return cc;
}
long long divi(long long a, long long b) { return a * power(b, MOD - 2) % MOD; }
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  getchar();
  for (int i = 0; i < (int)(k); i++) s[i] = getchar() == '+' ? 1 : -1;
  long long cur = 0LL;
  for (int i = 0; i < (int)(k); i++)
    cur = (cur + power(a, k - 1 - i) * power(b, i) * s[i]) % MOD;
  cur = (cur + MOD) % MOD;
  long long ca = power(a, k), cb = power(b, k);
  int ct = (n + 1) / k;
  if (ca == cb)
    cur *= power(cb, ct - 1) * ct;
  else
    cur *= power(cb, ct - 1) *
           divi(1 - power(divi(ca, cb), ct), 1 - divi(ca, cb)) % MOD;
  cur = (cur % MOD + MOD) % MOD;
  printf("%d\n", int(cur));
  return 0;
}

```