#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 9;
const double pi = acos(-1);
int n, t, i, j, k;
string s;
int a, b;
long long sum;
long long ans;
long long fastexp(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = fastexp(a, b / 2);
  temp = (1LL * temp * temp) % mod;
  if (b & 1) temp = (1LL * temp * a) % mod;
  return temp;
}
long long invers(long long a) {
  if (a == 1) return 1;
  long long ans = (1LL * -mod / a * invers(mod % a)) % mod;
  if (ans < 0) ans += mod;
  return ans;
}
int main() {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  cin >> s;
  sum = 0;
  long long ta = fastexp(a, k - 1);
  long long tb = 1;
  long long inva = invers(a);
  for (i = 0; i < k; i++) {
    int mul = s[i] == '+' ? 1 : -1;
    sum += (mul * ta * tb) % mod;
    sum %= mod;
    tb = (tb * b) % mod;
    ta = (ta * inva) % mod;
  }
  ans = 0;
  if (k != n + 1) {
    ans = fastexp(a, n + 1) - fastexp(b, n + 1);
    long long temp = fastexp(a, k) - fastexp(b, k);
    if (temp < 0) temp += mod;
    if (temp != 0) {
      temp = invers(temp);
      ans = (ans * temp) % mod;
    } else
      ans = (n + 1) / k;
  } else {
    ans = 1;
  }
  ans = (ans * sum) % mod;
  if (ans < 0) ans += 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;
const double pi = acos(-1);
int n, t, i, j, k;
string s;
int a, b;
long long sum;
long long ans;
long long fastexp(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = fastexp(a, b / 2);
  temp = (1LL * temp * temp) % mod;
  if (b & 1) temp = (1LL * temp * a) % mod;
  return temp;
}
long long invers(long long a) {
  if (a == 1) return 1;
  long long ans = (1LL * -mod / a * invers(mod % a)) % mod;
  if (ans < 0) ans += mod;
  return ans;
}
int main() {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  cin >> s;
  sum = 0;
  long long ta = fastexp(a, k - 1);
  long long tb = 1;
  long long inva = invers(a);
  for (i = 0; i < k; i++) {
    int mul = s[i] == '+' ? 1 : -1;
    sum += (mul * ta * tb) % mod;
    sum %= mod;
    tb = (tb * b) % mod;
    ta = (ta * inva) % mod;
  }
  ans = 0;
  if (k != n + 1) {
    ans = fastexp(a, n + 1) - fastexp(b, n + 1);
    long long temp = fastexp(a, k) - fastexp(b, k);
    if (temp < 0) temp += mod;
    if (temp != 0) {
      temp = invers(temp);
      ans = (ans * temp) % mod;
    } else
      ans = (n + 1) / k;
  } else {
    ans = 1;
  }
  ans = (ans * sum) % mod;
  if (ans < 0) ans += mod;
  printf("%lld\n", ans);
  return 0;
}

```