#include <bits/stdc++.h>
using namespace std;
bool vis[30000005], vis1[3000005];
vector<long long> vec[3000005];
long long final_sum[300005], max_sum[10005];
void dfs(long long node) {
  vis[node] = 1;
  for (long long i = 0; i < vec[node].size(); i++) {
    if (!vis[vec[node][i]]) {
      dfs(vec[node][i]);
      final_sum[node] += final_sum[vec[node][i]];
    }
  }
}
long long divi[10000000];
long long dp[1000007];
vector<long long> prime;
void run() {
  long long j = 1;
  for (long long i = 2; i < 1000001; i++) {
    if (!vis1[i])
      while (i * j < 1000001) {
        vis1[i * j] = 1;
        divi[i * j] = i;
        j++;
      }
    j = 1;
  }
  for (long long i = 2; i < 1000001; i++) {
    if (i == divi[i])
      dp[i] = dp[i - 1] + 1;
    else {
      dp[i] = dp[i - 1];
    }
  }
}
long long fastexpo(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % (long long)1000000009;
    b = b / 2;
    a = (a * a) % (long long)1000000009;
  }
  return ans % (long long)1000000009;
}
int32_t main() {
  long long a, n, b, k;
  cin >> n >> a >> b >> k;
  long long p = (n + 1) / k;
  string str;
  cin >> str;
  long long sum = 0;
  for (long long i = 0; i < k; i++) {
    if (str[i] == '+') {
      sum = (sum + fastexpo(a, k - i - 1) * fastexpo(b, i) +
             (long long)1000000009) %
            (long long)1000000009;
    } else {
      sum -= (fastexpo(a, k - i - 1) * fastexpo(b, i));
    }
    sum = (sum + 10000 * (long long)1000000009) % (long long)1000000009;
  }
  long long ans = (fastexpo(a, n + 1) - fastexpo(b, n + 1) +
                   10000 * (long long)1000000009) %
                  (long long)1000000009;
  long long q = (fastexpo(b, k) * fastexpo(a, k)) % (long long)1000000009;
  if (q == 1) {
    cout << (sum * p) % (long long)1000000009 << "\n";
    return 0;
  }
  long long ans1 =
      (fastexpo(a, k) - fastexpo(b, k) + 10000 * (long long)1000000009) %
      (long long)1000000009;
  ans1 = (fastexpo(ans1, (long long)1000000009 - 2)) % (long long)1000000009;
  ans = (ans * ans1) % (long long)1000000009;
  sum = (sum * ans) % (long long)1000000009;
  cout << sum << "\n";
  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>
using namespace std;
bool vis[30000005], vis1[3000005];
vector<long long> vec[3000005];
long long final_sum[300005], max_sum[10005];
void dfs(long long node) {
  vis[node] = 1;
  for (long long i = 0; i < vec[node].size(); i++) {
    if (!vis[vec[node][i]]) {
      dfs(vec[node][i]);
      final_sum[node] += final_sum[vec[node][i]];
    }
  }
}
long long divi[10000000];
long long dp[1000007];
vector<long long> prime;
void run() {
  long long j = 1;
  for (long long i = 2; i < 1000001; i++) {
    if (!vis1[i])
      while (i * j < 1000001) {
        vis1[i * j] = 1;
        divi[i * j] = i;
        j++;
      }
    j = 1;
  }
  for (long long i = 2; i < 1000001; i++) {
    if (i == divi[i])
      dp[i] = dp[i - 1] + 1;
    else {
      dp[i] = dp[i - 1];
    }
  }
}
long long fastexpo(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % (long long)1000000009;
    b = b / 2;
    a = (a * a) % (long long)1000000009;
  }
  return ans % (long long)1000000009;
}
int32_t main() {
  long long a, n, b, k;
  cin >> n >> a >> b >> k;
  long long p = (n + 1) / k;
  string str;
  cin >> str;
  long long sum = 0;
  for (long long i = 0; i < k; i++) {
    if (str[i] == '+') {
      sum = (sum + fastexpo(a, k - i - 1) * fastexpo(b, i) +
             (long long)1000000009) %
            (long long)1000000009;
    } else {
      sum -= (fastexpo(a, k - i - 1) * fastexpo(b, i));
    }
    sum = (sum + 10000 * (long long)1000000009) % (long long)1000000009;
  }
  long long ans = (fastexpo(a, n + 1) - fastexpo(b, n + 1) +
                   10000 * (long long)1000000009) %
                  (long long)1000000009;
  long long q = (fastexpo(b, k) * fastexpo(a, k)) % (long long)1000000009;
  if (q == 1) {
    cout << (sum * p) % (long long)1000000009 << "\n";
    return 0;
  }
  long long ans1 =
      (fastexpo(a, k) - fastexpo(b, k) + 10000 * (long long)1000000009) %
      (long long)1000000009;
  ans1 = (fastexpo(ans1, (long long)1000000009 - 2)) % (long long)1000000009;
  ans = (ans * ans1) % (long long)1000000009;
  sum = (sum * ans) % (long long)1000000009;
  cout << sum << "\n";
  return 0;
}

```