#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 1e9 + 9;
ll modPow(ll a, ll b) {
  if (b == 0) return 1;
  if (b & 1) return a * modPow(a, b - 1) % MOD;
  return modPow(a * a % MOD, b / 2);
}
ll geom(ll a, ll b) {
  if (a == 1)
    return b + 1;
  else if (a == 0)
    return 1;
  ll divi = (modPow(a, b + 1) + MOD - 1) % MOD;
  ll divs = modPow((a + MOD - 1) % MOD, MOD - 2);
  return divi * divs % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ll n, a, b, k;
  cin >> n >> a >> b >> k;
  ll t0 = modPow(a, n);
  a = modPow(a, MOD - 2);
  ll ab = a * b % MOD;
  ll mult = modPow(ab % MOD, k);
  ll sp = 0;
  string str;
  cin >> str;
  for (int i = 0; i < k; ++i) {
    if (str[i] == '+') {
      sp += t0;
    } else {
      sp += (MOD - t0) % MOD;
    }
    t0 = (t0 * ab) % MOD;
  }
  sp %= MOD;
  ll res = sp * geom(mult, (n + 1) / k - 1) % MOD;
  cout << res << '\n';
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const ll MOD = 1e9 + 9;
ll modPow(ll a, ll b) {
  if (b == 0) return 1;
  if (b & 1) return a * modPow(a, b - 1) % MOD;
  return modPow(a * a % MOD, b / 2);
}
ll geom(ll a, ll b) {
  if (a == 1)
    return b + 1;
  else if (a == 0)
    return 1;
  ll divi = (modPow(a, b + 1) + MOD - 1) % MOD;
  ll divs = modPow((a + MOD - 1) % MOD, MOD - 2);
  return divi * divs % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ll n, a, b, k;
  cin >> n >> a >> b >> k;
  ll t0 = modPow(a, n);
  a = modPow(a, MOD - 2);
  ll ab = a * b % MOD;
  ll mult = modPow(ab % MOD, k);
  ll sp = 0;
  string str;
  cin >> str;
  for (int i = 0; i < k; ++i) {
    if (str[i] == '+') {
      sp += t0;
    } else {
      sp += (MOD - t0) % MOD;
    }
    t0 = (t0 * ab) % MOD;
  }
  sp %= MOD;
  ll res = sp * geom(mult, (n + 1) / k - 1) % MOD;
  cout << res << '\n';
}

```