#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
const long long MOD = 1e9 + 9;
char s[MAX];
long long n, a, b, k;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long ret = power(x, y / 2);
  ret = (ret * ret) % MOD;
  if (y & 1) ret = (ret * x) % MOD;
  return ret;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long mult = (s[i] == '+') ? 1 : -1;
    long long ratio = (power(b, k) * power(power(a, k), MOD - 2)) % MOD;
    long long awal = (power(a, n - i) * power(b, i)) % MOD;
    long long sbnyk = (n + 1) / k - 1;
    if (ratio == 1) {
      ans = (ans + mult * awal * (sbnyk + 1)) % MOD;
      ans = (ans + MOD) % MOD;
    } else {
      long long bwh = power(ratio - 1, MOD - 2);
      long long atas = (awal * power(ratio, sbnyk + 1)) % MOD;
      atas -= awal;
      atas = (atas + MOD) % MOD;
      ans = (ans + mult * atas * bwh) % MOD;
      ans = (ans + MOD) % MOD;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MAX = 1e5 + 5;
const long long MOD = 1e9 + 9;
char s[MAX];
long long n, a, b, k;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long ret = power(x, y / 2);
  ret = (ret * ret) % MOD;
  if (y & 1) ret = (ret * x) % MOD;
  return ret;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    long long mult = (s[i] == '+') ? 1 : -1;
    long long ratio = (power(b, k) * power(power(a, k), MOD - 2)) % MOD;
    long long awal = (power(a, n - i) * power(b, i)) % MOD;
    long long sbnyk = (n + 1) / k - 1;
    if (ratio == 1) {
      ans = (ans + mult * awal * (sbnyk + 1)) % MOD;
      ans = (ans + MOD) % MOD;
    } else {
      long long bwh = power(ratio - 1, MOD - 2);
      long long atas = (awal * power(ratio, sbnyk + 1)) % MOD;
      atas -= awal;
      atas = (atas + MOD) % MOD;
      ans = (ans + mult * atas * bwh) % MOD;
      ans = (ans + MOD) % MOD;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```