#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k;
long long m = 1e9 + 9;
long long add(long long a, long long b) {
  return ((a % m + b % m) % m + m) % m;
}
long long mul(long long a, long long b) {
  return ((a % m * b % m) % m + m) % m;
}
const int sz = 1e5 + 9;
char s[sz];
long long base;
long long modpow(long long b, long long p) {
  if (b == 1) return 1;
  if (b == 0) return 0;
  if (p == 1) return b % m;
  if (p == 0) return 1;
  long long u = modpow(b, p >> 1);
  u = mul(u, u);
  if (p & 1) u = mul(u, b);
  return u;
}
int main() {
  scanf("%I64d %I64d %I64d %I64d", &n, &a, &b, &k);
  scanf(" %s", s);
  for (int i = 0; i < k; i++) {
    long long tmp = mul(modpow(a, n - i), modpow(b, i));
    tmp = mul(tmp, (s[i] == '+' ? 1 : -1));
    base = add(base, tmp);
  }
  long long cnt = (n + 1) / k;
  long long r = modpow(b, k);
  r = mul(r, modpow(modpow(a, k), m - 2));
  if (r == 1) {
    long long ans = mul(base, cnt);
    printf("%I64d", ans);
    return 0;
  }
  long long num = add(1, -modpow(r, cnt));
  long long den = modpow(add(1, -r), m - 2);
  long long ser = mul(num, den);
  long long ans = mul(base, ser);
  printf("%I64d", ans);
  return 0;
}

### Prompt

Generate 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;
long long n, a, b, k;
long long m = 1e9 + 9;
long long add(long long a, long long b) {
  return ((a % m + b % m) % m + m) % m;
}
long long mul(long long a, long long b) {
  return ((a % m * b % m) % m + m) % m;
}
const int sz = 1e5 + 9;
char s[sz];
long long base;
long long modpow(long long b, long long p) {
  if (b == 1) return 1;
  if (b == 0) return 0;
  if (p == 1) return b % m;
  if (p == 0) return 1;
  long long u = modpow(b, p >> 1);
  u = mul(u, u);
  if (p & 1) u = mul(u, b);
  return u;
}
int main() {
  scanf("%I64d %I64d %I64d %I64d", &n, &a, &b, &k);
  scanf(" %s", s);
  for (int i = 0; i < k; i++) {
    long long tmp = mul(modpow(a, n - i), modpow(b, i));
    tmp = mul(tmp, (s[i] == '+' ? 1 : -1));
    base = add(base, tmp);
  }
  long long cnt = (n + 1) / k;
  long long r = modpow(b, k);
  r = mul(r, modpow(modpow(a, k), m - 2));
  if (r == 1) {
    long long ans = mul(base, cnt);
    printf("%I64d", ans);
    return 0;
  }
  long long num = add(1, -modpow(r, cnt));
  long long den = modpow(add(1, -r), m - 2);
  long long ser = mul(num, den);
  long long ans = mul(base, ser);
  printf("%I64d", ans);
  return 0;
}

```