#include <bits/stdc++.h>
template <typename T>
T max(T x, T y) {
  return (x > y) ? x : y;
}
template <typename T>
T min(T x, T y) {
  return (x < y) ? x : y;
}
template <typename T>
bool chkmax(T &x, T y) {
  return (x >= y) ? 0 : (x = y, 1);
}
template <typename T>
bool chkmin(T &x, T y) {
  return (x <= y) ? 0 : (x = y, 1);
}
template <typename T>
T read(T &in) {
  in = 0;
  char ch;
  T f = 1;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) in = in * 10 + ch - '0', ch = getchar();
  return in *= f;
}
static const int MOD = 1000000009;
static const int max1 = 10000000 + 11;
int add(int x) { return (x >= MOD) ? (x - MOD) : x; }
int reduce(int x) { return (x < 0) ? (x + MOD) : x; }
void Add(int &x) {
  if (x >= MOD) x -= MOD;
}
void Reduce(int &x) {
  if (x < 0) x += MOD;
}
int qpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % MOD)
    if (y & 1) res = 1ll * res * x % MOD;
  return res;
}
int inv(int x) { return qpow(x, MOD - 2); }
int n, a, b, c, k;
char S[max1];
int calc(int c, int n) {
  if (c == 1) return n + 1;
  int fac1 = reduce(qpow(c, n + 1) - 1);
  int fac2 = inv(reduce(c - 1));
  return 1ll * fac1 * fac2 % MOD;
}
int main() {
  read(n), read(a), read(b), read(k);
  scanf("%s", S);
  c = 1ll * b * inv(a) % MOD;
  int Ans = 0, ck = qpow(c, k), _c = 1;
  for (int i = 0; i < k; ++i, _c = 1ll * _c * c % MOD) {
    int s = (S[i] == '+') ? 1 : -1;
    int now = 1ll * calc(ck, (n - i) / k) * _c % MOD;
    if (s == 1)
      Add(Ans += now);
    else
      Reduce(Ans -= now);
  }
  Ans = 1ll * Ans * qpow(a, n) % MOD;
  printf("%d\n", Ans);
  return 0;
}

### Prompt

Your task is to create 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>
template <typename T>
T max(T x, T y) {
  return (x > y) ? x : y;
}
template <typename T>
T min(T x, T y) {
  return (x < y) ? x : y;
}
template <typename T>
bool chkmax(T &x, T y) {
  return (x >= y) ? 0 : (x = y, 1);
}
template <typename T>
bool chkmin(T &x, T y) {
  return (x <= y) ? 0 : (x = y, 1);
}
template <typename T>
T read(T &in) {
  in = 0;
  char ch;
  T f = 1;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) in = in * 10 + ch - '0', ch = getchar();
  return in *= f;
}
static const int MOD = 1000000009;
static const int max1 = 10000000 + 11;
int add(int x) { return (x >= MOD) ? (x - MOD) : x; }
int reduce(int x) { return (x < 0) ? (x + MOD) : x; }
void Add(int &x) {
  if (x >= MOD) x -= MOD;
}
void Reduce(int &x) {
  if (x < 0) x += MOD;
}
int qpow(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, x = 1ll * x * x % MOD)
    if (y & 1) res = 1ll * res * x % MOD;
  return res;
}
int inv(int x) { return qpow(x, MOD - 2); }
int n, a, b, c, k;
char S[max1];
int calc(int c, int n) {
  if (c == 1) return n + 1;
  int fac1 = reduce(qpow(c, n + 1) - 1);
  int fac2 = inv(reduce(c - 1));
  return 1ll * fac1 * fac2 % MOD;
}
int main() {
  read(n), read(a), read(b), read(k);
  scanf("%s", S);
  c = 1ll * b * inv(a) % MOD;
  int Ans = 0, ck = qpow(c, k), _c = 1;
  for (int i = 0; i < k; ++i, _c = 1ll * _c * c % MOD) {
    int s = (S[i] == '+') ? 1 : -1;
    int now = 1ll * calc(ck, (n - i) / k) * _c % MOD;
    if (s == 1)
      Add(Ans += now);
    else
      Reduce(Ans -= now);
  }
  Ans = 1ll * Ans * qpow(a, n) % MOD;
  printf("%d\n", Ans);
  return 0;
}

```