#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000009;
long long modpow(long long x, long long n, long long m) {
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
int main() {
  int t = 1, co = 1;
  while (t--) {
    long long n, d, i, j, k, a, b, q, z = 0, ans = 0;
    cin >> n >> a >> b >> k;
    string s;
    cin >> s;
    for (i = 0; i < k; i++) {
      if (s[i] == '+')
        z += modpow(a, n - i, mod) * modpow(b, i, mod);
      else
        z -= modpow(a, n - i, mod) * modpow(b, i, mod);
      z = (z + mod) % mod;
    }
    q = (b * modpow(a, mod - 2, mod)) % mod;
    q = modpow(q, k, mod);
    long long gp;
    if (q == 1) {
      gp = (n + 1) / k;
    } else {
      gp = modpow(q, (n + 1) / k, mod) - 1;
      if (gp < 0) gp += mod;
      j = modpow(q - 1, mod - 2, mod);
      gp = (gp * j) % mod;
    }
    cout << ((z * gp) % mod + 10 * mod) % mod;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long mod = 1000000009;
long long modpow(long long x, long long n, long long m) {
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
int main() {
  int t = 1, co = 1;
  while (t--) {
    long long n, d, i, j, k, a, b, q, z = 0, ans = 0;
    cin >> n >> a >> b >> k;
    string s;
    cin >> s;
    for (i = 0; i < k; i++) {
      if (s[i] == '+')
        z += modpow(a, n - i, mod) * modpow(b, i, mod);
      else
        z -= modpow(a, n - i, mod) * modpow(b, i, mod);
      z = (z + mod) % mod;
    }
    q = (b * modpow(a, mod - 2, mod)) % mod;
    q = modpow(q, k, mod);
    long long gp;
    if (q == 1) {
      gp = (n + 1) / k;
    } else {
      gp = modpow(q, (n + 1) / k, mod) - 1;
      if (gp < 0) gp += mod;
      j = modpow(q - 1, mod - 2, mod);
      gp = (gp * j) % mod;
    }
    cout << ((z * gp) % mod + 10 * mod) % mod;
  }
  return 0;
}

```