#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
ll modularExponentiation(ll x, ll n, ll mod) {
  x %= mod;
  ll y = 1;
  for (; n > 0; n >>= 1, x = (x * x) % mod)
    if (n & 1) y = (y * x) % mod;
  return y;
}
const ll MOD = 9 + (ll)1e9;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  ll n, a, b, k;
  cin >> n >> a >> b >> k;
  char sg[k];
  cin >> sg;
  ll terms = (n + 1) / k;
  ll r = (b * modularExponentiation(a, MOD - 2, MOD)) % MOD;
  ll rk = modularExponentiation(r, k, MOD);
  ll base = modularExponentiation(a, n, MOD);
  if (rk != 1) {
    base =
        (base * (modularExponentiation(rk, terms, MOD) + MOD - 1) % MOD) % MOD;
    base = (base * modularExponentiation((rk + MOD - 1) % MOD, MOD - 2, MOD)) %
           MOD;
  } else {
    base = (base * terms) % MOD;
  }
  ll tot = 0;
  for (int i = 0; i < k; i++) {
    if (sg[i] == '+')
      tot = (tot + base) % MOD;
    else
      tot = (tot + MOD - base) % MOD;
    base = (base * r) % MOD;
  }
  cout << tot << endl;
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
using ld = long double;
ll modularExponentiation(ll x, ll n, ll mod) {
  x %= mod;
  ll y = 1;
  for (; n > 0; n >>= 1, x = (x * x) % mod)
    if (n & 1) y = (y * x) % mod;
  return y;
}
const ll MOD = 9 + (ll)1e9;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand(time(0));
  ll n, a, b, k;
  cin >> n >> a >> b >> k;
  char sg[k];
  cin >> sg;
  ll terms = (n + 1) / k;
  ll r = (b * modularExponentiation(a, MOD - 2, MOD)) % MOD;
  ll rk = modularExponentiation(r, k, MOD);
  ll base = modularExponentiation(a, n, MOD);
  if (rk != 1) {
    base =
        (base * (modularExponentiation(rk, terms, MOD) + MOD - 1) % MOD) % MOD;
    base = (base * modularExponentiation((rk + MOD - 1) % MOD, MOD - 2, MOD)) %
           MOD;
  } else {
    base = (base * terms) % MOD;
  }
  ll tot = 0;
  for (int i = 0; i < k; i++) {
    if (sg[i] == '+')
      tot = (tot + base) % MOD;
    else
      tot = (tot + MOD - base) % MOD;
    base = (base * r) % MOD;
  }
  cout << tot << endl;
  return 0;
}

```