#include <bits/stdc++.h>
const int maxn = 1e5 + 10;
const long long mod = 1000000009;
using namespace std;
char s[maxn];
long long n, a, b, k;
long long quickmod(long long x, long long y) {
  long long t = x % mod;
  long long ans = 1;
  while (y) {
    if (y % 2 == 1) {
      ans = (ans * t) % mod;
    }
    y /= 2;
    t = (t * t) % mod;
  }
  return ans;
}
long long solution() {
  long long ans = 0;
  long long nn = k - 1;
  for (long long i = 0; i <= nn; i++) {
    if (s[i] == '-') {
      ans = (ans - (quickmod(a, nn - i) * quickmod(b, i)) % mod + mod) % mod;
    } else if (s[i] == '+') {
      ans = (ans + (quickmod(a, nn - i) * quickmod(b, i)) % mod) % mod;
    }
  }
  return ans;
}
long long solve(long long x) {
  long long ans = 0;
  for (long long i = n - x + 1; i <= n; i++) {
    if (s[i] == '-') {
      ans = (ans - (quickmod(a, n - i) * quickmod(b, i)) % mod + mod) % mod;
    } else if (s[i] == '+') {
      ans = (ans + (quickmod(a, n - i) * quickmod(b, i)) % mod) % mod;
    }
  }
  return ans;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long m = (n + 1) / k;
  long long y = (n + 1) % k;
  long long d = solution();
  long long qq = quickmod(a, mod - 2);
  long long q = quickmod((qq * b) % mod, k);
  long long aa = quickmod(a, n - k + 1);
  long long ans = 0;
  if (q % mod == 1) {
    ans = (((aa * m) % mod) * d) % mod;
  } else {
    long long aaa = quickmod(1 - q, mod - 2);
    ans = (((aa * aaa) % mod) * ((1 - quickmod(q, m) + mod) % mod)) % mod;
    ans = ((ans * d) % mod + mod) % mod;
  }
  if (y == 0) {
    cout << ans << endl;
  } else {
    ans = ((ans * quickmod(a, y)) % mod + solve(y) + mod) % mod;
    cout << ans << endl;
  }
  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>
const int maxn = 1e5 + 10;
const long long mod = 1000000009;
using namespace std;
char s[maxn];
long long n, a, b, k;
long long quickmod(long long x, long long y) {
  long long t = x % mod;
  long long ans = 1;
  while (y) {
    if (y % 2 == 1) {
      ans = (ans * t) % mod;
    }
    y /= 2;
    t = (t * t) % mod;
  }
  return ans;
}
long long solution() {
  long long ans = 0;
  long long nn = k - 1;
  for (long long i = 0; i <= nn; i++) {
    if (s[i] == '-') {
      ans = (ans - (quickmod(a, nn - i) * quickmod(b, i)) % mod + mod) % mod;
    } else if (s[i] == '+') {
      ans = (ans + (quickmod(a, nn - i) * quickmod(b, i)) % mod) % mod;
    }
  }
  return ans;
}
long long solve(long long x) {
  long long ans = 0;
  for (long long i = n - x + 1; i <= n; i++) {
    if (s[i] == '-') {
      ans = (ans - (quickmod(a, n - i) * quickmod(b, i)) % mod + mod) % mod;
    } else if (s[i] == '+') {
      ans = (ans + (quickmod(a, n - i) * quickmod(b, i)) % mod) % mod;
    }
  }
  return ans;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long m = (n + 1) / k;
  long long y = (n + 1) % k;
  long long d = solution();
  long long qq = quickmod(a, mod - 2);
  long long q = quickmod((qq * b) % mod, k);
  long long aa = quickmod(a, n - k + 1);
  long long ans = 0;
  if (q % mod == 1) {
    ans = (((aa * m) % mod) * d) % mod;
  } else {
    long long aaa = quickmod(1 - q, mod - 2);
    ans = (((aa * aaa) % mod) * ((1 - quickmod(q, m) + mod) % mod)) % mod;
    ans = ((ans * d) % mod + mod) % mod;
  }
  if (y == 0) {
    cout << ans << endl;
  } else {
    ans = ((ans * quickmod(a, y)) % mod + solve(y) + mod) % mod;
    cout << ans << endl;
  }
  return 0;
}

```