#include <bits/stdc++.h>
using namespace std;
const long long N = 100100;
const long long Mod = 1e9 + 9;
char s[N];
long long add(long long x, long long y) { return (x + y + 10 * Mod) % Mod; }
void add_self(long long& x, long long y) { x = add(x, y); }
long long sub(long long x, long long y) { return (x - y + 10 * Mod) % Mod; }
void sub_self(long long& x, long long y) { x = sub(x, y); }
long long mul(long long x, long long y) { return x * y % Mod; }
void mul_self(long long& x, long long y) { x = mul(x, y); }
long long fp(long long x, long long y) {
  if (!y) return 1;
  long long Res = fp(x, y >> 1);
  mul_self(Res, Res);
  if (y & 1) mul_self(Res, x);
  return Res;
}
long long inv(long long y) { return fp(y, Mod - 2); }
long long inv(long long x, long long y) { return mul(x, inv(y)); }
long long Sum(long long q, long long m) {
  if (q == 1) return m;
  return inv(sub(fp(q, m), 1), sub(q, 1));
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  long long A = fp(a, n);
  long long q = inv(b, a);
  long long base = fp(q, k);
  long long m = (n + 1) / k;
  long long Ans = 0;
  for (long long i = 0; i < k; i++) {
    long long staff = fp(q, i);
    long long whatever = mul(staff, Sum(base, m));
    s[i] == '+' ? add_self(Ans, whatever) : sub_self(Ans, whatever);
  }
  mul_self(Ans, A);
  cout << Ans;
}

### Prompt

Please create a solution in CPP 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 long long N = 100100;
const long long Mod = 1e9 + 9;
char s[N];
long long add(long long x, long long y) { return (x + y + 10 * Mod) % Mod; }
void add_self(long long& x, long long y) { x = add(x, y); }
long long sub(long long x, long long y) { return (x - y + 10 * Mod) % Mod; }
void sub_self(long long& x, long long y) { x = sub(x, y); }
long long mul(long long x, long long y) { return x * y % Mod; }
void mul_self(long long& x, long long y) { x = mul(x, y); }
long long fp(long long x, long long y) {
  if (!y) return 1;
  long long Res = fp(x, y >> 1);
  mul_self(Res, Res);
  if (y & 1) mul_self(Res, x);
  return Res;
}
long long inv(long long y) { return fp(y, Mod - 2); }
long long inv(long long x, long long y) { return mul(x, inv(y)); }
long long Sum(long long q, long long m) {
  if (q == 1) return m;
  return inv(sub(fp(q, m), 1), sub(q, 1));
}
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  long long A = fp(a, n);
  long long q = inv(b, a);
  long long base = fp(q, k);
  long long m = (n + 1) / k;
  long long Ans = 0;
  for (long long i = 0; i < k; i++) {
    long long staff = fp(q, i);
    long long whatever = mul(staff, Sum(base, m));
    s[i] == '+' ? add_self(Ans, whatever) : sub_self(Ans, whatever);
  }
  mul_self(Ans, A);
  cout << Ans;
}

```