#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 9;
long long n, a, b, k;
char str[100005];
long long fe(long long x, long long p) {
  if (p == 0) return 1LL;
  if (p == 1) return x;
  long long ret = fe(x, p / 2);
  ret = (ret * ret) % MOD;
  if (p % 2 == 1) {
    ret = (ret * x) % MOD;
  }
  return ret;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  getchar();
  scanf("%s", str);
  int pan = strlen(str);
  long long cur = 1LL;
  long long ans = 0LL;
  for (int i = 0; i < pan; i++) {
    if (i == 0)
      cur = 1LL;
    else
      cur = (cur * b) % MOD;
    long long tmp = (fe(a, n - i) * cur) % MOD;
    if (str[i] == '+')
      ans += tmp;
    else
      ans -= tmp;
    while (ans < 0) ans += MOD;
    ans %= MOD;
  }
  long long ul = (n + 1) / k;
  long long ra = fe(a, k);
  ra = fe(ra, MOD - 2);
  ra = (ra * fe(b, k)) % MOD;
  if (ra == 1) {
    printf("%lld\n", (ans * ul) % MOD);
    return 0;
  }
  long long ka = fe(ra, ul);
  ka--;
  if (ka < 0) ka += MOD;
  ka %= MOD;
  long long baw = fe(ra - 1, MOD - 2);
  ans = (ans * ka + MOD) % MOD;
  ans = (ans * baw + MOD) % MOD;
  printf("%lld\n", ans);
  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>
using namespace std;
const long long MOD = 1e9 + 9;
long long n, a, b, k;
char str[100005];
long long fe(long long x, long long p) {
  if (p == 0) return 1LL;
  if (p == 1) return x;
  long long ret = fe(x, p / 2);
  ret = (ret * ret) % MOD;
  if (p % 2 == 1) {
    ret = (ret * x) % MOD;
  }
  return ret;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  getchar();
  scanf("%s", str);
  int pan = strlen(str);
  long long cur = 1LL;
  long long ans = 0LL;
  for (int i = 0; i < pan; i++) {
    if (i == 0)
      cur = 1LL;
    else
      cur = (cur * b) % MOD;
    long long tmp = (fe(a, n - i) * cur) % MOD;
    if (str[i] == '+')
      ans += tmp;
    else
      ans -= tmp;
    while (ans < 0) ans += MOD;
    ans %= MOD;
  }
  long long ul = (n + 1) / k;
  long long ra = fe(a, k);
  ra = fe(ra, MOD - 2);
  ra = (ra * fe(b, k)) % MOD;
  if (ra == 1) {
    printf("%lld\n", (ans * ul) % MOD);
    return 0;
  }
  long long ka = fe(ra, ul);
  ka--;
  if (ka < 0) ka += MOD;
  ka %= MOD;
  long long baw = fe(ra - 1, MOD - 2);
  ans = (ans * ka + MOD) % MOD;
  ans = (ans * baw + MOD) % MOD;
  printf("%lld\n", ans);
  return 0;
}

```