#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const double PI = 3.14159265359;
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() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int mod = 1e9 + 9;
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  vector<int> v(k);
  for (int& i : v) {
    char c;
    cin >> c;
    i = c == '+' ? 1 : -1;
  }
  long long z = 0;
  for (int i = 0; i < k; i++) {
    z += v[i] * modpow(a, n - i, mod) * modpow(b, i, mod);
    if (z < mod) z += mod;
    z %= mod;
  }
  long long ainv = modpow(a, mod - 2, mod);
  auto q = modpow(ainv * b % mod, k, mod);
  if (q == 1) {
    long long ans = z * (n + 1) / k;
    cout << ans % mod;
    return 0;
  }
  long long num = 1 - modpow(q, (n + 1) / k, mod);
  num += mod;
  num %= mod;
  auto denom = 1 - q;
  denom += mod;
  denom %= mod;
  auto denominv = modpow(denom, mod - 2, mod);
  long long ans = num * denominv;
  ans %= mod;
  ans *= z;
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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 MOD = 1e9 + 7;
const double PI = 3.14159265359;
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() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int mod = 1e9 + 9;
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  vector<int> v(k);
  for (int& i : v) {
    char c;
    cin >> c;
    i = c == '+' ? 1 : -1;
  }
  long long z = 0;
  for (int i = 0; i < k; i++) {
    z += v[i] * modpow(a, n - i, mod) * modpow(b, i, mod);
    if (z < mod) z += mod;
    z %= mod;
  }
  long long ainv = modpow(a, mod - 2, mod);
  auto q = modpow(ainv * b % mod, k, mod);
  if (q == 1) {
    long long ans = z * (n + 1) / k;
    cout << ans % mod;
    return 0;
  }
  long long num = 1 - modpow(q, (n + 1) / k, mod);
  num += mod;
  num %= mod;
  auto denom = 1 - q;
  denom += mod;
  denom %= mod;
  auto denominv = modpow(denom, mod - 2, mod);
  long long ans = num * denominv;
  ans %= mod;
  ans *= z;
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans;
  return 0;
}

```