#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int inf = 0x3f3f3f3f;
const double e = 1e-8;
const long long mod = 1e9 + 9;
long long qPow(long long a, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n % 2 == 1) ans *= a % mod;
    ans = ans % mod;
    a *= a % mod;
    a = a % mod;
    n /= 2;
  }
  return ans % mod;
}
char s[maxn];
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  getchar();
  for (int i = 0; i < k; i++) {
    scanf("%c", &s[i]);
  }
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      ans += qPow(a, n - i) * qPow(b, i) % mod;
    } else {
      ans -= (qPow(a, n - i) * qPow(b, i) + mod) % mod;
    }
    ans = (ans + mod) % mod;
  }
  long long f = (n + 1) / k;
  long long bkf = qPow(b, k * f) % mod;
  long long akf = qPow(a, k * f) % mod;
  long long ak_f = qPow(a, k * f - k) % mod;
  long long bk = qPow(b, k) % mod;
  long long ak = qPow(a, k) % mod;
  long long mak = qPow(ak, mod - 2) % mod;
  long long ff = (bk * mak) % mod;
  if (ff == 1)
    ans = (ans * f) % mod;
  else {
    long long di = ((bk - ak + mod) % mod * ak_f % mod + mod) % mod;
    long long sha = (bkf - akf + mod) % mod;
    long long niyuan = qPow(di, mod - 2) % mod;
    ans = (ans * sha % mod);
    ans *= niyuan % mod;
    ans = ans % mod;
  }
  cout << ans << endl;
}

### 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 int maxn = 1e5 + 5;
const int inf = 0x3f3f3f3f;
const double e = 1e-8;
const long long mod = 1e9 + 9;
long long qPow(long long a, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n % 2 == 1) ans *= a % mod;
    ans = ans % mod;
    a *= a % mod;
    a = a % mod;
    n /= 2;
  }
  return ans % mod;
}
char s[maxn];
int main() {
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  getchar();
  for (int i = 0; i < k; i++) {
    scanf("%c", &s[i]);
  }
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+') {
      ans += qPow(a, n - i) * qPow(b, i) % mod;
    } else {
      ans -= (qPow(a, n - i) * qPow(b, i) + mod) % mod;
    }
    ans = (ans + mod) % mod;
  }
  long long f = (n + 1) / k;
  long long bkf = qPow(b, k * f) % mod;
  long long akf = qPow(a, k * f) % mod;
  long long ak_f = qPow(a, k * f - k) % mod;
  long long bk = qPow(b, k) % mod;
  long long ak = qPow(a, k) % mod;
  long long mak = qPow(ak, mod - 2) % mod;
  long long ff = (bk * mak) % mod;
  if (ff == 1)
    ans = (ans * f) % mod;
  else {
    long long di = ((bk - ak + mod) % mod * ak_f % mod + mod) % mod;
    long long sha = (bkf - akf + mod) % mod;
    long long niyuan = qPow(di, mod - 2) % mod;
    ans = (ans * sha % mod);
    ans *= niyuan % mod;
    ans = ans % mod;
  }
  cout << ans << endl;
}

```