#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int M = 100005;
const long long MOD = 1e9 + 9;
const double pi = acos(-1);
long long pow_mod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
long long get_inv(long long a) { return pow_mod(a, MOD - 2); }
char s[M];
int main() {
  long long n, a, b, k, ans = 0;
  cin >> n >> a >> b >> k;
  getchar();
  cin.getline(s, k + 1);
  long long t = n / k + 1, inv1 = get_inv(a),
            inv2 = get_inv(pow_mod(inv1 * 1ll * b % MOD, k) - 1);
  for (int i = 0; i < k; ++i) {
    long long tp = pow_mod(a, n - i) * 1ll * pow_mod(b, i) % MOD;
    if (inv2)
      tp = tp * 1ll * (pow_mod(pow_mod(inv1 * 1ll * b % MOD, k), t) - 1) % MOD *
           inv2 % MOD;
    else
      tp = tp * t % MOD;
    if (s[i] == '-') tp = MOD - tp;
    ans = (ans + tp) % MOD;
  }
  cout << ans << endl;
  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 int INF = 0x3f3f3f3f;
const int M = 100005;
const long long MOD = 1e9 + 9;
const double pi = acos(-1);
long long pow_mod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
long long get_inv(long long a) { return pow_mod(a, MOD - 2); }
char s[M];
int main() {
  long long n, a, b, k, ans = 0;
  cin >> n >> a >> b >> k;
  getchar();
  cin.getline(s, k + 1);
  long long t = n / k + 1, inv1 = get_inv(a),
            inv2 = get_inv(pow_mod(inv1 * 1ll * b % MOD, k) - 1);
  for (int i = 0; i < k; ++i) {
    long long tp = pow_mod(a, n - i) * 1ll * pow_mod(b, i) % MOD;
    if (inv2)
      tp = tp * 1ll * (pow_mod(pow_mod(inv1 * 1ll * b % MOD, k), t) - 1) % MOD *
           inv2 % MOD;
    else
      tp = tp * t % MOD;
    if (s[i] == '-') tp = MOD - tp;
    ans = (ans + tp) % MOD;
  }
  cout << ans << endl;
  return 0;
}

```