#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
const long long p = (long long)1e9 + 9;
long long Pow(long long a, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 0) {
    return Pow((a * a) % p, n / 2);
  }
  return (Pow((a * a) % p, n / 2) * a) % p;
}
map<int, long long> dp;
long long DP(long long c, long long d, int m) {
  if (m == 0) {
    return 1;
  }
  if (m == 1) {
    return (c + d) % p;
  }
  if (dp.find(m) != dp.end()) {
    return dp[m];
  }
  int mid_m = (m - 1) / 2;
  return dp[m] = (Pow(c, m - mid_m) * DP(c, d, mid_m) +
                  Pow(d, mid_m + 1) * DP(c, d, m - mid_m - 1)) %
                 p;
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  vector<long long> a_pow(k + 1), b_pow(k + 1);
  a_pow[0] = 1, b_pow[0] = 1;
  for (int i = 1; i <= k; i++) {
    a_pow[i] = (a * a_pow[i - 1]) % p;
    b_pow[i] = (b * b_pow[i - 1]) % p;
  }
  long long ans = 0;
  for (int i = 0; i < (k); i++) {
    char a;
    cin >> a;
    long long sign;
    if (a == '+') {
      sign = 1;
    } else {
      sign = -1;
    }
    ans = (ans + sign * a_pow[k - 1 - i] * b_pow[i]) % p;
  }
  ans *= DP(a_pow[k], b_pow[k], (n - k + 1) / k);
  ans %= p;
  ans += p;
  ans %= p;
  cout << ans << endl;
}

### 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>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
const long long p = (long long)1e9 + 9;
long long Pow(long long a, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 0) {
    return Pow((a * a) % p, n / 2);
  }
  return (Pow((a * a) % p, n / 2) * a) % p;
}
map<int, long long> dp;
long long DP(long long c, long long d, int m) {
  if (m == 0) {
    return 1;
  }
  if (m == 1) {
    return (c + d) % p;
  }
  if (dp.find(m) != dp.end()) {
    return dp[m];
  }
  int mid_m = (m - 1) / 2;
  return dp[m] = (Pow(c, m - mid_m) * DP(c, d, mid_m) +
                  Pow(d, mid_m + 1) * DP(c, d, m - mid_m - 1)) %
                 p;
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  long long n, a, b, k;
  cin >> n >> a >> b >> k;
  vector<long long> a_pow(k + 1), b_pow(k + 1);
  a_pow[0] = 1, b_pow[0] = 1;
  for (int i = 1; i <= k; i++) {
    a_pow[i] = (a * a_pow[i - 1]) % p;
    b_pow[i] = (b * b_pow[i - 1]) % p;
  }
  long long ans = 0;
  for (int i = 0; i < (k); i++) {
    char a;
    cin >> a;
    long long sign;
    if (a == '+') {
      sign = 1;
    } else {
      sign = -1;
    }
    ans = (ans + sign * a_pow[k - 1 - i] * b_pow[i]) % p;
  }
  ans *= DP(a_pow[k], b_pow[k], (n - k + 1) / k);
  ans %= p;
  ans += p;
  ans %= p;
  cout << ans << endl;
}

```