#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 9;
int pow1(int a, int b) {
  int ret = 1;
  int temp = a;
  while (b) {
    if (b & 1) ret = 1LL * ret * temp % MOD;
    temp = 1LL * temp % MOD * temp % MOD;
    b >>= 1;
  }
  return ret;
}
long long cir = 0, ans = 0, dech = 0;
int main() {
  int n, a, b, k, tt;
  char c;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  getchar();
  for (int i = 0; i < k; ++i) {
    c = getchar();
    tt = 1LL * pow1(a, n - i) * pow1(b, i) % MOD;
    tt = c == '+' ? tt : -tt;
    cir = (cir + tt + MOD) % MOD;
  }
  tt = (n + 1) / k;
  long long up, down, res, invd;
  up = 1LL * (pow1(a, n + 1) - pow1(b, n + 1) + MOD) % MOD;
  down = 1LL * (pow1(a, n + 1) - 1LL * pow1(b, k) * pow1(a, n + 1 - k) + MOD) %
         MOD;
  invd = pow1(down, MOD - 2);
  if (up == 0) {
    ans = 1LL * cir * tt % MOD;
  } else {
    ans = 1LL * (cir * up % MOD * invd % MOD) % MOD;
  }
  cout << (ans + MOD) % MOD << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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 MOD = 1e9 + 9;
int pow1(int a, int b) {
  int ret = 1;
  int temp = a;
  while (b) {
    if (b & 1) ret = 1LL * ret * temp % MOD;
    temp = 1LL * temp % MOD * temp % MOD;
    b >>= 1;
  }
  return ret;
}
long long cir = 0, ans = 0, dech = 0;
int main() {
  int n, a, b, k, tt;
  char c;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  getchar();
  for (int i = 0; i < k; ++i) {
    c = getchar();
    tt = 1LL * pow1(a, n - i) * pow1(b, i) % MOD;
    tt = c == '+' ? tt : -tt;
    cir = (cir + tt + MOD) % MOD;
  }
  tt = (n + 1) / k;
  long long up, down, res, invd;
  up = 1LL * (pow1(a, n + 1) - pow1(b, n + 1) + MOD) % MOD;
  down = 1LL * (pow1(a, n + 1) - 1LL * pow1(b, k) * pow1(a, n + 1 - k) + MOD) %
         MOD;
  invd = pow1(down, MOD - 2);
  if (up == 0) {
    ans = 1LL * cir * tt % MOD;
  } else {
    ans = 1LL * (cir * up % MOD * invd % MOD) % MOD;
  }
  cout << (ans + MOD) % MOD << endl;
  return 0;
}

```