#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 9;
char st[1000005];
int x[1000005];
inline int read() {
  char ch = getchar(), last = '!';
  int ans = 0;
  while (ch < '0' || ch > '9') last = ch, ch = getchar();
  while (ch <= '9' && ch >= '0') ans = ans * 10 + ch - 48, ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % P;
    a = (a * a) % P;
    b >>= 1;
  }
  return ans;
}
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", st + 1);
  for (int i = 1; i <= k; i++)
    if (st[i] == '+')
      x[i] = 1;
    else
      x[i] = -1;
  long long t = (n + 1) / k, ans = 0;
  long long a1 = 1, xa = ksm(a, P - 2);
  long long q = ksm(xa, k) * ksm(b, k) % P;
  long long x1 = ((1 - ksm(q, t)) % P + P) % P, x2 = ksm(1 - q, P - 2);
  for (int i = 1; i <= k; i++) {
    a1 = a1 * xa % P * b % P;
    if (i == 1) a1 = 1;
    if (q == 1)
      ans += t * a1 % P * x[i];
    else
      ans += a1 * x1 % P * x2 % P * x[i];
    ans = (ans % P + P) % P;
  }
  printf("%I64d\n", ans * ksm(a, n) % P);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 P = 1e9 + 9;
char st[1000005];
int x[1000005];
inline int read() {
  char ch = getchar(), last = '!';
  int ans = 0;
  while (ch < '0' || ch > '9') last = ch, ch = getchar();
  while (ch <= '9' && ch >= '0') ans = ans * 10 + ch - 48, ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % P;
    a = (a * a) % P;
    b >>= 1;
  }
  return ans;
}
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", st + 1);
  for (int i = 1; i <= k; i++)
    if (st[i] == '+')
      x[i] = 1;
    else
      x[i] = -1;
  long long t = (n + 1) / k, ans = 0;
  long long a1 = 1, xa = ksm(a, P - 2);
  long long q = ksm(xa, k) * ksm(b, k) % P;
  long long x1 = ((1 - ksm(q, t)) % P + P) % P, x2 = ksm(1 - q, P - 2);
  for (int i = 1; i <= k; i++) {
    a1 = a1 * xa % P * b % P;
    if (i == 1) a1 = 1;
    if (q == 1)
      ans += t * a1 % P * x[i];
    else
      ans += a1 * x1 % P * x2 % P * x[i];
    ans = (ans % P + P) % P;
  }
  printf("%I64d\n", ans * ksm(a, n) % P);
  return 0;
}

```