#include <bits/stdc++.h>
using namespace std;
long long int russian_peasant(long long int a, long long int b) {
  long long int res = 0;
  while (a > 0) {
    if (a & 1) {
      res += b;
      if (res >= 1000000009LL) res %= 1000000009LL;
    }
    a >>= 1;
    b <<= 1;
    if (b >= 1000000009LL) b %= 1000000009LL;
  }
  return res;
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % 1000000009LL;
  while (y > 0) {
    if (y & 1) res = russian_peasant(res, x) % 1000000009LL;
    y = y >> 1;
    x = russian_peasant(x, x) % 1000000009LL;
  }
  return res;
}
long long int inverse_modulo(long long int x) {
  return power(x, 1000000009LL - 2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  long long int store_a[100100], store_b[100100];
  string s;
  cin >> s;
  store_a[0] = power(a, n - k + 1);
  store_b[0] = 1;
  for (int i = 0; i < k; i++) {
    if (i > 0) {
      store_a[i] = russian_peasant(store_a[i - 1], a);
      store_b[i] = russian_peasant(store_b[i - 1], b);
    }
  }
  long long int Z = 0;
  for (int i = 0; i < k; i++) {
    long long int help = russian_peasant(store_a[k - 1 - i], store_b[i]);
    help = help * ((s[i] == '+') ? (1) : (-1));
    Z += help;
    Z %= 1000000009LL;
  }
  long long int Q = power(russian_peasant(b, inverse_modulo(a)), k);
  Z = (Z + 1000000009LL) % 1000000009LL;
  if (Q == 1) {
    Z = russian_peasant(Z, (n + 1) / k);
  } else {
    long long int h =
        russian_peasant(power(Q, (n + 1) / k) - 1, inverse_modulo(Q - 1));
    h = (h + 1000000009LL) % 1000000009LL;
    Z = russian_peasant(Z, h);
  }
  cout << (Z + 1000000009LL) % 1000000009LL << 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>
using namespace std;
long long int russian_peasant(long long int a, long long int b) {
  long long int res = 0;
  while (a > 0) {
    if (a & 1) {
      res += b;
      if (res >= 1000000009LL) res %= 1000000009LL;
    }
    a >>= 1;
    b <<= 1;
    if (b >= 1000000009LL) b %= 1000000009LL;
  }
  return res;
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % 1000000009LL;
  while (y > 0) {
    if (y & 1) res = russian_peasant(res, x) % 1000000009LL;
    y = y >> 1;
    x = russian_peasant(x, x) % 1000000009LL;
  }
  return res;
}
long long int inverse_modulo(long long int x) {
  return power(x, 1000000009LL - 2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  long long int store_a[100100], store_b[100100];
  string s;
  cin >> s;
  store_a[0] = power(a, n - k + 1);
  store_b[0] = 1;
  for (int i = 0; i < k; i++) {
    if (i > 0) {
      store_a[i] = russian_peasant(store_a[i - 1], a);
      store_b[i] = russian_peasant(store_b[i - 1], b);
    }
  }
  long long int Z = 0;
  for (int i = 0; i < k; i++) {
    long long int help = russian_peasant(store_a[k - 1 - i], store_b[i]);
    help = help * ((s[i] == '+') ? (1) : (-1));
    Z += help;
    Z %= 1000000009LL;
  }
  long long int Q = power(russian_peasant(b, inverse_modulo(a)), k);
  Z = (Z + 1000000009LL) % 1000000009LL;
  if (Q == 1) {
    Z = russian_peasant(Z, (n + 1) / k);
  } else {
    long long int h =
        russian_peasant(power(Q, (n + 1) / k) - 1, inverse_modulo(Q - 1));
    h = (h + 1000000009LL) % 1000000009LL;
    Z = russian_peasant(Z, h);
  }
  cout << (Z + 1000000009LL) % 1000000009LL << endl;
  return 0;
}

```