#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <int MOD>
struct mod_int {
  static const int Mod = MOD;
  unsigned x;
  mod_int() : x(0) {}
  mod_int(int sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  mod_int(long long sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  int get() const { return (int)x; }
  mod_int &operator+=(mod_int that) {
    if ((x += that.x) >= MOD) x -= MOD;
    return *this;
  }
  mod_int &operator-=(mod_int that) {
    if ((x += MOD - that.x) >= MOD) x -= MOD;
    return *this;
  }
  mod_int &operator*=(mod_int that) {
    x = (unsigned long long)x * that.x % MOD;
    return *this;
  }
  mod_int &operator/=(mod_int that) { return *this *= that.inverse(); }
  mod_int operator+(mod_int that) const { return mod_int(*this) += that; }
  mod_int operator-(mod_int that) const { return mod_int(*this) -= that; }
  mod_int operator*(mod_int that) const { return mod_int(*this) *= that; }
  mod_int operator/(mod_int that) const { return mod_int(*this) /= that; }
  mod_int inverse() const {
    long long a = x, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    return mod_int(u);
  }
};
using mint = mod_int<1000000009>;
mint power(mint a, ll b) {
  mint res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  ll n, a, b, k;
  string s;
  cin >> n >> a >> b >> k >> s;
  if (n < k) {
    mint res = 0;
    for (ll i = 0; i <= n; i++) {
      if (s[i] == '+') {
        res += power(a, n - i) * power(b, i);
      } else {
        res -= power(a, n - i) * power(b, i);
      }
    }
    cout << res.get() << endl;
    return 0;
  }
  mint res = 0;
  mint x = power(mint(b) / mint(a), k);
  if (x.get() == 1) {
    for (ll i = 0; i < k; i++) {
      ll len = (n + 1 - i + k - 1) / k;
      if (s[i] == '+') {
        res += power(a, n - i) * power(b, i) * len;
      } else {
        res -= power(a, n - i) * power(b, i) * len;
      }
    }
  } else {
    for (ll i = 0; i < k; i++) {
      ll len = (n + 1 - i + k - 1) / k;
      if (s[i] == '+') {
        res += power(a, n - i) * power(b, i) *
               ((mint(1) - power(x, len)) / (mint(1) - x));
      } else {
        res -= power(a, n - i) * power(b, i) *
               ((mint(1) - power(x, len)) / (mint(1) - x));
      }
    }
  }
  cout << res.get() << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
using ll = long long;
template <int MOD>
struct mod_int {
  static const int Mod = MOD;
  unsigned x;
  mod_int() : x(0) {}
  mod_int(int sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  mod_int(long long sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  int get() const { return (int)x; }
  mod_int &operator+=(mod_int that) {
    if ((x += that.x) >= MOD) x -= MOD;
    return *this;
  }
  mod_int &operator-=(mod_int that) {
    if ((x += MOD - that.x) >= MOD) x -= MOD;
    return *this;
  }
  mod_int &operator*=(mod_int that) {
    x = (unsigned long long)x * that.x % MOD;
    return *this;
  }
  mod_int &operator/=(mod_int that) { return *this *= that.inverse(); }
  mod_int operator+(mod_int that) const { return mod_int(*this) += that; }
  mod_int operator-(mod_int that) const { return mod_int(*this) -= that; }
  mod_int operator*(mod_int that) const { return mod_int(*this) *= that; }
  mod_int operator/(mod_int that) const { return mod_int(*this) /= that; }
  mod_int inverse() const {
    long long a = x, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    return mod_int(u);
  }
};
using mint = mod_int<1000000009>;
mint power(mint a, ll b) {
  mint res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  ll n, a, b, k;
  string s;
  cin >> n >> a >> b >> k >> s;
  if (n < k) {
    mint res = 0;
    for (ll i = 0; i <= n; i++) {
      if (s[i] == '+') {
        res += power(a, n - i) * power(b, i);
      } else {
        res -= power(a, n - i) * power(b, i);
      }
    }
    cout << res.get() << endl;
    return 0;
  }
  mint res = 0;
  mint x = power(mint(b) / mint(a), k);
  if (x.get() == 1) {
    for (ll i = 0; i < k; i++) {
      ll len = (n + 1 - i + k - 1) / k;
      if (s[i] == '+') {
        res += power(a, n - i) * power(b, i) * len;
      } else {
        res -= power(a, n - i) * power(b, i) * len;
      }
    }
  } else {
    for (ll i = 0; i < k; i++) {
      ll len = (n + 1 - i + k - 1) / k;
      if (s[i] == '+') {
        res += power(a, n - i) * power(b, i) *
               ((mint(1) - power(x, len)) / (mint(1) - x));
      } else {
        res -= power(a, n - i) * power(b, i) *
               ((mint(1) - power(x, len)) / (mint(1) - x));
      }
    }
  }
  cout << res.get() << endl;
  return 0;
}

```