#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x) {
  x = 0;
  bool f = 0;
  char ch = (char)getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = (char)getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch & 15);
    ch = (char)getchar();
  }
  if (f) x = -x;
}
template <typename T>
inline void write(T x) {
  static int st[35];
  int top = 0;
  do {
    st[top++] = x % 10;
    x /= 10;
  } while (x);
  while (top) putchar(st[--top] + '0');
}
const double pi = acos(-1.0);
const double eps = (double)1e-4;
const int mod = 1e9 + 9;
const int maxn = 1e5 + 5;
const int maxm = 2e5 + 10;
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) (res *= a) %= mod;
    (a *= a) %= mod;
    b >>= 1;
  }
  return res;
}
long long inv(long long a) { return qpow(a, mod - 2); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long inva = inv(a), invb = inv(b);
  long long ak = inv(qpow(a, k)), bk = qpow(b, k);
  long long q = ak * bk % mod;
  long long d = (n + 1) / k;
  string s;
  cin >> s;
  long long cura = qpow(a, n), curb = 1;
  long long base = 0;
  for (int i = 0; i < s.size(); i++) {
    int f;
    if (s[i] == '+')
      f = 1;
    else
      f = -1;
    base = (base + (long long)(f) * (cura * curb % mod)) % mod;
    cura = cura * inva % mod;
    curb = curb * b % mod;
  }
  long long ans;
  if (q == 1)
    ans = (d * base % mod + mod) % mod;
  else {
    long long de = (qpow(q, d) - 1LL) % mod;
    ans = de * inv(q - 1) % mod * base % mod;
    ans = (ans + mod) % mod;
  }
  cout << ans << endl;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
template <typename T>
inline void read(T& x) {
  x = 0;
  bool f = 0;
  char ch = (char)getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = (char)getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch & 15);
    ch = (char)getchar();
  }
  if (f) x = -x;
}
template <typename T>
inline void write(T x) {
  static int st[35];
  int top = 0;
  do {
    st[top++] = x % 10;
    x /= 10;
  } while (x);
  while (top) putchar(st[--top] + '0');
}
const double pi = acos(-1.0);
const double eps = (double)1e-4;
const int mod = 1e9 + 9;
const int maxn = 1e5 + 5;
const int maxm = 2e5 + 10;
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) (res *= a) %= mod;
    (a *= a) %= mod;
    b >>= 1;
  }
  return res;
}
long long inv(long long a) { return qpow(a, mod - 2); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  long long inva = inv(a), invb = inv(b);
  long long ak = inv(qpow(a, k)), bk = qpow(b, k);
  long long q = ak * bk % mod;
  long long d = (n + 1) / k;
  string s;
  cin >> s;
  long long cura = qpow(a, n), curb = 1;
  long long base = 0;
  for (int i = 0; i < s.size(); i++) {
    int f;
    if (s[i] == '+')
      f = 1;
    else
      f = -1;
    base = (base + (long long)(f) * (cura * curb % mod)) % mod;
    cura = cura * inva % mod;
    curb = curb * b % mod;
  }
  long long ans;
  if (q == 1)
    ans = (d * base % mod + mod) % mod;
  else {
    long long de = (qpow(q, d) - 1LL) % mod;
    ans = de * inv(q - 1) % mod * base % mod;
    ans = (ans + mod) % mod;
  }
  cout << ans << endl;
}

```