#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 9;
inline void fix(long long &x) {
  x %= mod;
  x += mod;
  x %= mod;
}
inline long long power(long long a, long long k) {
  long long ans = 1;
  fix(a);
  while (k) {
    if (k & 1) {
      ans = ans * a % mod;
    }
    a = a * a % mod;
    k >>= 1;
  }
  fix(ans);
  return ans;
}
long long N, A, B, K, fuck;
inline long long inv(long long x) {
  fix(x);
  return power(x, mod - 2);
}
inline long long mul(long long a, long long b) {
  fix(a);
  fix(b);
  return a * b % mod;
}
const int N_MAX = 1e5 + 10;
inline long long shit() {
  if (power(fuck, K) == 1 || fuck == 1 || A == B) {
    return (N + 1) / K;
  } else {
    return mul(1 - power(fuck, N + 1), inv(1 - power(fuck, K)));
  }
}
char s[N_MAX];
long long sgn[N_MAX];
int main() {
  scanf("%lld%lld%lld%lld", &N, &A, &B, &K);
  scanf("%s", s + 1);
  for (int i = 1; i <= K; i++) {
    if (s[i] == '+') {
      sgn[i - 1] = 1;
    } else {
      sgn[i - 1] = -1;
    }
  }
  long long ans = 0;
  fuck = mul(B, inv(A));
  for (int k = 0; k < K; k++) {
    ans += sgn[k] * (power(fuck, k));
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  ans = mul(ans, shit());
  ans = mul(ans, power(A, N));
  ans = (ans + mod) % mod;
  printf("%lld\n", ans);
}

### Prompt

Develop a solution in Cpp 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 = 1e9 + 9;
inline void fix(long long &x) {
  x %= mod;
  x += mod;
  x %= mod;
}
inline long long power(long long a, long long k) {
  long long ans = 1;
  fix(a);
  while (k) {
    if (k & 1) {
      ans = ans * a % mod;
    }
    a = a * a % mod;
    k >>= 1;
  }
  fix(ans);
  return ans;
}
long long N, A, B, K, fuck;
inline long long inv(long long x) {
  fix(x);
  return power(x, mod - 2);
}
inline long long mul(long long a, long long b) {
  fix(a);
  fix(b);
  return a * b % mod;
}
const int N_MAX = 1e5 + 10;
inline long long shit() {
  if (power(fuck, K) == 1 || fuck == 1 || A == B) {
    return (N + 1) / K;
  } else {
    return mul(1 - power(fuck, N + 1), inv(1 - power(fuck, K)));
  }
}
char s[N_MAX];
long long sgn[N_MAX];
int main() {
  scanf("%lld%lld%lld%lld", &N, &A, &B, &K);
  scanf("%s", s + 1);
  for (int i = 1; i <= K; i++) {
    if (s[i] == '+') {
      sgn[i - 1] = 1;
    } else {
      sgn[i - 1] = -1;
    }
  }
  long long ans = 0;
  fuck = mul(B, inv(A));
  for (int k = 0; k < K; k++) {
    ans += sgn[k] * (power(fuck, k));
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  ans = mul(ans, shit());
  ans = mul(ans, power(A, N));
  ans = (ans + mod) % mod;
  printf("%lld\n", ans);
}

```