#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const double PI = 4 * atan(1);
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 9;
const int MAX_N = 2e6 + 10;
ll n, a, b, k;
string s;
ll powmod(ll x, ll y) {
  ll res = 1;
  while (y) {
    if (y & 1) (res *= x) %= MOD;
    (x *= x) %= MOD;
    y >>= 1;
  }
  return res;
}
void solve() {
  ll z = 0;
  for (int i = 0; i < k; ++i) {
    ll c = powmod(a, n - i) * powmod(b, i) % MOD;
    if (s[i] == '-') c = MOD - c;
    z += c;
  }
  z %= MOD;
  ll q = powmod(b * powmod(a, MOD - 2) % MOD, k);
  ll nk = (n + 1) / k;
  if (q == 1) {
    z *= nk;
  } else {
    z *= (powmod(q, nk) - 1 + MOD) % MOD * powmod(q - 1, MOD - 2) % MOD;
  }
  cout << z % MOD << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> a >> b >> k;
  cin >> s;
  solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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 pii = pair<int, int>;
const double PI = 4 * atan(1);
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 9;
const int MAX_N = 2e6 + 10;
ll n, a, b, k;
string s;
ll powmod(ll x, ll y) {
  ll res = 1;
  while (y) {
    if (y & 1) (res *= x) %= MOD;
    (x *= x) %= MOD;
    y >>= 1;
  }
  return res;
}
void solve() {
  ll z = 0;
  for (int i = 0; i < k; ++i) {
    ll c = powmod(a, n - i) * powmod(b, i) % MOD;
    if (s[i] == '-') c = MOD - c;
    z += c;
  }
  z %= MOD;
  ll q = powmod(b * powmod(a, MOD - 2) % MOD, k);
  ll nk = (n + 1) / k;
  if (q == 1) {
    z *= nk;
  } else {
    z *= (powmod(q, nk) - 1 + MOD) % MOD * powmod(q - 1, MOD - 2) % MOD;
  }
  cout << z % MOD << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> a >> b >> k;
  cin >> s;
  solve();
  return 0;
}

```