#include <bits/stdc++.h>
using namespace std;
const int MOD = 1E9 + 9;
inline int Add(int a, int b) {
  return (a + b >= MOD) ? (a + b - MOD) : (a + b);
}
inline int Sub(int a, int b) { return (a >= b) ? (a - b) : (a - b + MOD); }
inline int Mul(int a, int b) { return (long long)a * b % MOD; }
inline int Pow(int a, int b) {
  int c = 1;
  for (; b; a = Mul(a, a), b >>= 1)
    if (b & 1) c = Mul(c, a);
  return c;
}
inline int Div(int a, int b) { return Mul(a, Pow(b, MOD - 2)); }
const int N = 100005;
int n, a, b, k;
char ch[N];
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", ch);
  int ret = 0;
  for (int i = 0; i < k; ++i) {
    int t = Mul(Pow(a, n - i), Pow(b, i));
    if (ch[i] == '+')
      ret = Add(ret, t);
    else
      ret = Sub(ret, t);
  }
  int x = Mul(Pow(Pow(a, MOD - 2), k), Pow(b, k));
  n = (n + 1) / k;
  if (x == 1)
    ret = Mul(ret, n);
  else
    ret = Mul(ret, Div(Sub(Pow(x, n), 1), Sub(x, 1)));
  printf("%d\n", ret);
  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;
inline int Add(int a, int b) {
  return (a + b >= MOD) ? (a + b - MOD) : (a + b);
}
inline int Sub(int a, int b) { return (a >= b) ? (a - b) : (a - b + MOD); }
inline int Mul(int a, int b) { return (long long)a * b % MOD; }
inline int Pow(int a, int b) {
  int c = 1;
  for (; b; a = Mul(a, a), b >>= 1)
    if (b & 1) c = Mul(c, a);
  return c;
}
inline int Div(int a, int b) { return Mul(a, Pow(b, MOD - 2)); }
const int N = 100005;
int n, a, b, k;
char ch[N];
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", ch);
  int ret = 0;
  for (int i = 0; i < k; ++i) {
    int t = Mul(Pow(a, n - i), Pow(b, i));
    if (ch[i] == '+')
      ret = Add(ret, t);
    else
      ret = Sub(ret, t);
  }
  int x = Mul(Pow(Pow(a, MOD - 2), k), Pow(b, k));
  n = (n + 1) / k;
  if (x == 1)
    ret = Mul(ret, n);
  else
    ret = Mul(ret, Div(Sub(Pow(x, n), 1), Sub(x, 1)));
  printf("%d\n", ret);
  return 0;
}

```