#include <bits/stdc++.h>
const long long MOD = 1000000009;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x * 10) + (ch ^ 48);
  return x * f;
}
inline long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1LL) res = res * a % MOD;
    b >>= 1, a = a * a % MOD;
  }
  return res;
}
long long n, a, b, inva, ans;
int k;
char s[100010];
int main() {
  int i;
  n = read(), a = read(), b = read(), k = read();
  inva = qpow(a, MOD - 2);
  scanf("%s", s);
  long long val = qpow(a, n);
  for (i = 0; i < k; i++) {
    if (s[i] == '-')
      ans = (ans + MOD - val) % MOD;
    else
      ans = (ans + val) % MOD;
    val = val * inva % MOD * b % MOD;
  }
  val = qpow(b * inva % MOD, k);
  if (val == 1)
    ans = ans * ((n + 1) / k) % MOD;
  else {
    ans = ans * qpow((val + MOD - 1) % MOD, MOD - 2) % MOD;
    ans = ans * ((qpow(inva * b % MOD, n + 1) % MOD + MOD - 1) % MOD) % MOD;
  }
  std::cout << ans << std::endl;
  return 0;
}

### Prompt

Please formulate 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>
const long long MOD = 1000000009;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x * 10) + (ch ^ 48);
  return x * f;
}
inline long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1LL) res = res * a % MOD;
    b >>= 1, a = a * a % MOD;
  }
  return res;
}
long long n, a, b, inva, ans;
int k;
char s[100010];
int main() {
  int i;
  n = read(), a = read(), b = read(), k = read();
  inva = qpow(a, MOD - 2);
  scanf("%s", s);
  long long val = qpow(a, n);
  for (i = 0; i < k; i++) {
    if (s[i] == '-')
      ans = (ans + MOD - val) % MOD;
    else
      ans = (ans + val) % MOD;
    val = val * inva % MOD * b % MOD;
  }
  val = qpow(b * inva % MOD, k);
  if (val == 1)
    ans = ans * ((n + 1) / k) % MOD;
  else {
    ans = ans * qpow((val + MOD - 1) % MOD, MOD - 2) % MOD;
    ans = ans * ((qpow(inva * b % MOD, n + 1) % MOD + MOD - 1) % MOD) % MOD;
  }
  std::cout << ans << std::endl;
  return 0;
}

```