#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 9;
long long int power(long long int x, long long int y) {
  long long int ans = 1;
  x = x % MOD;
  while (y > 0) {
    if (y & 1) {
      ans *= x;
      ans %= MOD;
    }
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return ans;
}
long long int inverseMod(long long int a) { return power(a, MOD - 2); }
void solve() {
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  string st;
  cin >> st;
  long long int ans = 0;
  for (long long int i = 0; i < k; i++) {
    if (st[i] == '+') {
      ans += (power(a, n - i) * power(b, i)) % MOD;
    } else {
      ans -= (power(a, n - i) * power(b, i)) % MOD;
    }
    ans += MOD;
    ans %= MOD;
  }
  long long int factor = power(b, k);
  long long int p = power(a, k);
  factor *= inverseMod(p);
  factor %= MOD;
  long long int ch = (n + 1) / k;
  if (factor == 1) {
    ans *= ch;
    ans %= MOD;
    cout << ans << endl;
  } else {
    long long int val = power(factor, ch);
    val--;
    val += MOD;
    val %= MOD;
    factor--;
    factor += MOD;
    factor %= MOD;
    factor = inverseMod(factor);
    ans *= val;
    ans %= MOD;
    ans *= factor;
    ans %= MOD;
    cout << ans << endl;
  }
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
const long long int MOD = 1e9 + 9;
long long int power(long long int x, long long int y) {
  long long int ans = 1;
  x = x % MOD;
  while (y > 0) {
    if (y & 1) {
      ans *= x;
      ans %= MOD;
    }
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return ans;
}
long long int inverseMod(long long int a) { return power(a, MOD - 2); }
void solve() {
  long long int n, a, b, k;
  cin >> n >> a >> b >> k;
  string st;
  cin >> st;
  long long int ans = 0;
  for (long long int i = 0; i < k; i++) {
    if (st[i] == '+') {
      ans += (power(a, n - i) * power(b, i)) % MOD;
    } else {
      ans -= (power(a, n - i) * power(b, i)) % MOD;
    }
    ans += MOD;
    ans %= MOD;
  }
  long long int factor = power(b, k);
  long long int p = power(a, k);
  factor *= inverseMod(p);
  factor %= MOD;
  long long int ch = (n + 1) / k;
  if (factor == 1) {
    ans *= ch;
    ans %= MOD;
    cout << ans << endl;
  } else {
    long long int val = power(factor, ch);
    val--;
    val += MOD;
    val %= MOD;
    factor--;
    factor += MOD;
    factor %= MOD;
    factor = inverseMod(factor);
    ans *= val;
    ans %= MOD;
    ans *= factor;
    ans %= MOD;
    cout << ans << endl;
  }
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```