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

### Prompt

Please formulate 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 = 1000000009;
const int MAXK = 100005;
int n, a, b, k;
char s[MAXK];
int power(int a, int n) {
  int ans = 1;
  while (n) {
    if (n & 1) {
      ans = (long long)ans * a % MOD;
    }
    a = (long long)a * a % MOD, n >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", s);
  int ans = 0;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+') {
      int r = (long long)power(a, MOD - 1 - k) * power(b, k) % MOD;
      int x = (long long)power(a, n - i) * power(b, i) % MOD;
      if (r == 1) {
        ans = (ans + x * (n / k + 1ll)) % MOD;
        continue;
      }
      ans = (ans + x * (1ll - power(r, n / k + 1) + MOD) % MOD *
                       power(1 - r + MOD, MOD - 2)) %
            MOD;
    } else {
      int r = (long long)power(a, MOD - 1 - k) * power(b, k) % MOD;
      int x = (MOD - 1ll) * power(a, n - i) % MOD * power(b, i) % MOD;
      ans = (ans + x * (1ll - power(r, n / k + 1) + MOD) % MOD *
                       power(1 - r + MOD, MOD - 2)) %
            MOD;
      if (r == 1) {
        ans = (ans + x * (n / k + 1ll)) % MOD;
        continue;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```