#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
constexpr int N = 100005;
constexpr int MOD = 1000000009;
ll res[N];
int sizes[N];
ll pow(ll val, ll p, ll mod) {
  ll res = 1;
  while (p) {
    if (p & 1) res = (res * val) % mod;
    val = (val * val) % mod;
    p >>= 1;
  }
  return res;
}
ll reversed(ll val, ll mod) { return pow(val, mod - 2, mod); }
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, a, b, k;
  string s;
  cin >> n >> a >> b >> k >> s;
  int rev_a = reversed(a, MOD);
  for (int i = 0; i < k; i++) {
    res[0] = (res[0] + (s[i] == '+' ? 1 : MOD - 1) * pow(a, n - i, MOD) % MOD *
                           pow(b, i, MOD) % MOD) %
             MOD;
  }
  sizes[0] = k;
  int mx = 0;
  for (int i = 1; sizes[i - 1] * 2 <= n + 1; i++) {
    res[i] = (res[i - 1] + res[i - 1] * pow(rev_a, sizes[i - 1], MOD) % MOD *
                               pow(b, sizes[i - 1], MOD) % MOD) %
             MOD;
    sizes[i] = sizes[i - 1] * 2;
    mx = i;
  }
  int done = 0;
  ll ans = 0;
  for (int i = mx; i >= 0; i--) {
    if (sizes[i] <= n + 1 - done) {
      ans = (ans +
             res[i] * pow(rev_a, done, MOD) % MOD * pow(b, done, MOD) % MOD) %
            MOD;
      done += sizes[i];
    }
  }
  for (int i = done; i <= n; i++) {
    ans = (ans + (s[i % k] == '+' ? 1 : MOD - 1) * pow(a, n - i, MOD) % MOD *
                     pow(b, i, MOD) % MOD) %
          MOD;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
constexpr int N = 100005;
constexpr int MOD = 1000000009;
ll res[N];
int sizes[N];
ll pow(ll val, ll p, ll mod) {
  ll res = 1;
  while (p) {
    if (p & 1) res = (res * val) % mod;
    val = (val * val) % mod;
    p >>= 1;
  }
  return res;
}
ll reversed(ll val, ll mod) { return pow(val, mod - 2, mod); }
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, a, b, k;
  string s;
  cin >> n >> a >> b >> k >> s;
  int rev_a = reversed(a, MOD);
  for (int i = 0; i < k; i++) {
    res[0] = (res[0] + (s[i] == '+' ? 1 : MOD - 1) * pow(a, n - i, MOD) % MOD *
                           pow(b, i, MOD) % MOD) %
             MOD;
  }
  sizes[0] = k;
  int mx = 0;
  for (int i = 1; sizes[i - 1] * 2 <= n + 1; i++) {
    res[i] = (res[i - 1] + res[i - 1] * pow(rev_a, sizes[i - 1], MOD) % MOD *
                               pow(b, sizes[i - 1], MOD) % MOD) %
             MOD;
    sizes[i] = sizes[i - 1] * 2;
    mx = i;
  }
  int done = 0;
  ll ans = 0;
  for (int i = mx; i >= 0; i--) {
    if (sizes[i] <= n + 1 - done) {
      ans = (ans +
             res[i] * pow(rev_a, done, MOD) % MOD * pow(b, done, MOD) % MOD) %
            MOD;
      done += sizes[i];
    }
  }
  for (int i = done; i <= n; i++) {
    ans = (ans + (s[i % k] == '+' ? 1 : MOD - 1) * pow(a, n - i, MOD) % MOD *
                     pow(b, i, MOD) % MOD) %
          MOD;
  }
  cout << ans << endl;
  return 0;
}

```