#include <bits/stdc++.h>
using namespace std;
const int MOD = 1E9 + 9;
int Add(int a, int b, int m = MOD) {
  a += b;
  if (a >= m) a -= m;
  return a;
}
int Sub(int a, int b, int m = MOD) {
  a -= b;
  if (a < 0) a += m;
  return a;
}
int AddTo(int& to, int w, int m = MOD) {
  to += w;
  if (to >= m) to -= m;
  return to;
}
int SubTo(int& to, int w, int m = MOD) {
  to -= w;
  if (to < 0) to += m;
  return to;
}
int Mult(int a, int b, int m = MOD) { return (a * 1LL * b) % m; }
int Pow(int a, long long deg, int m = MOD) {
  int res = 1 % m;
  for (; deg > 0; deg >>= 1, a = Mult(a, a, m))
    if (deg & 1) res = Mult(res, a, m);
  return res;
}
int Oem(int to, int m = MOD) { return Pow(to, m - 2, m); }
int Div(int a, int b, int m = MOD) { return Mult(a, Oem(b, m), m); }
int n, a, b, k;
char buf[100010];
int SumProg(int how, int koef) {
  if (how == 0) return 0;
  if (how == 1) return 1;
  if (how % 2 == 1) {
    return Add(SumProg(how - 1, koef), Pow(koef, how - 1));
  } else {
    int s = SumProg(how / 2, koef);
    return Add(s, Mult(s, Pow(koef, how / 2)));
  }
}
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", buf);
  int base = 0;
  for (int i = 0; i < k; ++i) {
    if (buf[i] == '-') {
      SubTo(base, Mult(Pow(a, n - i), Pow(b, i)));
    } else {
      AddTo(base, Mult(Pow(a, n - i), Pow(b, i)));
    }
  }
  int koef = Div(Pow(b, k), Pow(a, k));
  int sum = SumProg((n + 1) / k, koef);
  printf("%d\n", Mult(sum, base));
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int Add(int a, int b, int m = MOD) {
  a += b;
  if (a >= m) a -= m;
  return a;
}
int Sub(int a, int b, int m = MOD) {
  a -= b;
  if (a < 0) a += m;
  return a;
}
int AddTo(int& to, int w, int m = MOD) {
  to += w;
  if (to >= m) to -= m;
  return to;
}
int SubTo(int& to, int w, int m = MOD) {
  to -= w;
  if (to < 0) to += m;
  return to;
}
int Mult(int a, int b, int m = MOD) { return (a * 1LL * b) % m; }
int Pow(int a, long long deg, int m = MOD) {
  int res = 1 % m;
  for (; deg > 0; deg >>= 1, a = Mult(a, a, m))
    if (deg & 1) res = Mult(res, a, m);
  return res;
}
int Oem(int to, int m = MOD) { return Pow(to, m - 2, m); }
int Div(int a, int b, int m = MOD) { return Mult(a, Oem(b, m), m); }
int n, a, b, k;
char buf[100010];
int SumProg(int how, int koef) {
  if (how == 0) return 0;
  if (how == 1) return 1;
  if (how % 2 == 1) {
    return Add(SumProg(how - 1, koef), Pow(koef, how - 1));
  } else {
    int s = SumProg(how / 2, koef);
    return Add(s, Mult(s, Pow(koef, how / 2)));
  }
}
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", buf);
  int base = 0;
  for (int i = 0; i < k; ++i) {
    if (buf[i] == '-') {
      SubTo(base, Mult(Pow(a, n - i), Pow(b, i)));
    } else {
      AddTo(base, Mult(Pow(a, n - i), Pow(b, i)));
    }
  }
  int koef = Div(Pow(b, k), Pow(a, k));
  int sum = SumProg((n + 1) / k, koef);
  printf("%d\n", Mult(sum, base));
  return 0;
}

```