#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = (1e9) + 9;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y % 2) {
    return ((x % mod) * (power(x, y - 1) % mod)) % mod;
  } else
    return (power((x * x) % mod, y / 2) % mod);
}
int n, a, b, k;
int main() {
  scanf("%d %d", &n, &a);
  scanf("%d %d", &b, &k);
  char temp[100005];
  scanf("%s", temp);
  long long x = 0;
  for (int i = 0; i < k; ++i) {
    if (temp[i] == '+')
      x = (x + (power(a, k - 1 - i) * power(b, i) % mod)) % mod;
    else
      x = (x - (power(a, k - 1 - i) * power(b, i) % mod)) % mod;
  }
  int m = (n + 1) / k;
  long long alp = power(a, k), bet = power(b, k), ans = 0;
  if (alp - bet) {
    ans = (((x * power(alp - bet, mod - 2)) % mod) *
           ((power(alp, m) - power(bet, m)) % mod)) %
          mod;
  } else {
    ans = (((x * m) % mod) * power(alp, m - 1)) % mod;
  }
  if (ans < 0)
    printf("%lld\n", mod + ans);
  else
    printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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 INF = 0x3f3f3f3f;
const int mod = (1e9) + 9;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y % 2) {
    return ((x % mod) * (power(x, y - 1) % mod)) % mod;
  } else
    return (power((x * x) % mod, y / 2) % mod);
}
int n, a, b, k;
int main() {
  scanf("%d %d", &n, &a);
  scanf("%d %d", &b, &k);
  char temp[100005];
  scanf("%s", temp);
  long long x = 0;
  for (int i = 0; i < k; ++i) {
    if (temp[i] == '+')
      x = (x + (power(a, k - 1 - i) * power(b, i) % mod)) % mod;
    else
      x = (x - (power(a, k - 1 - i) * power(b, i) % mod)) % mod;
  }
  int m = (n + 1) / k;
  long long alp = power(a, k), bet = power(b, k), ans = 0;
  if (alp - bet) {
    ans = (((x * power(alp - bet, mod - 2)) % mod) *
           ((power(alp, m) - power(bet, m)) % mod)) %
          mod;
  } else {
    ans = (((x * m) % mod) * power(alp, m - 1)) % mod;
  }
  if (ans < 0)
    printf("%lld\n", mod + ans);
  else
    printf("%lld\n", ans);
  return 0;
}

```