#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000009;
long long ex(long long x, long long y) {
  if (y == 0) return 1;
  long long r = ex(x, y >> 1);
  r *= r;
  r %= MOD;
  if (y & 1) {
    r *= x;
    r %= MOD;
  }
  return r;
}
char buf[100005];
long long n, a, b, ans;
int k;
int main() {
  scanf("%lld%lld%lld%d%s", &n, &a, &b, &k, buf);
  for (int i = 0; i < k; i++) {
    long long here = MOD - 1;
    if (buf[i] == '+') here = 1;
    here *= ex(a, n - i);
    here %= MOD;
    here *= ex(b, i);
    here %= MOD;
    ans += here;
    ans %= MOD;
  }
  long long rat = ex(a, MOD - 1 - k) * ex(b, k);
  rat %= MOD;
  if (rat == 1) {
    long long fact = (n + 1) / k;
    ans = ans * fact;
    ans %= MOD;
  } else {
    rat = (1 - ex(rat, (n + 1) / k) + MOD) * ex(1 - rat + MOD, MOD - 2);
    rat %= MOD;
    ans = (ans * rat) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}

### 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 long long MOD = 1000000009;
long long ex(long long x, long long y) {
  if (y == 0) return 1;
  long long r = ex(x, y >> 1);
  r *= r;
  r %= MOD;
  if (y & 1) {
    r *= x;
    r %= MOD;
  }
  return r;
}
char buf[100005];
long long n, a, b, ans;
int k;
int main() {
  scanf("%lld%lld%lld%d%s", &n, &a, &b, &k, buf);
  for (int i = 0; i < k; i++) {
    long long here = MOD - 1;
    if (buf[i] == '+') here = 1;
    here *= ex(a, n - i);
    here %= MOD;
    here *= ex(b, i);
    here %= MOD;
    ans += here;
    ans %= MOD;
  }
  long long rat = ex(a, MOD - 1 - k) * ex(b, k);
  rat %= MOD;
  if (rat == 1) {
    long long fact = (n + 1) / k;
    ans = ans * fact;
    ans %= MOD;
  } else {
    rat = (1 - ex(rat, (n + 1) / k) + MOD) * ex(1 - rat + MOD, MOD - 2);
    rat %= MOD;
    ans = (ans * rat) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}

```