#include <bits/stdc++.h>
using namespace std;
using LL = int64_t;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 9;
const int INF = 0x3f3f3f3f;
const LL LLINF = 1e14;
const double pi = acos(-1.0);
char s[maxn];
LL q_pow(LL x, LL n) {
  LL ans = 1;
  x %= mod;
  while (n) {
    if (n & 1) ans = (ans * x) % mod;
    n >>= 1;
    x = (x * x) % mod;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  LL n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> s;
  LL ans = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      ans = (ans + q_pow(a, n - i) % mod * q_pow(b, i) % mod + mod) % mod;
    else
      ans = (ans - q_pow(a, n - i) % mod * q_pow(b, i) % mod + mod) % mod;
  }
  LL t = (n + 1) / k;
  LL q = q_pow(a, k * (mod - 2) % (mod - 1)) % mod * q_pow(b, k) % mod;
  if (q == 1)
    ans = (ans * t + mod) % mod;
  else
    ans = (ans * (1 - q_pow(q, t)) % mod * q_pow(1 - q, mod - 2) % mod + mod) %
          mod;
  if (ans < 0) ans += mod;
  cout << ans;
  return 0;
}

### Prompt

Your task is to create a CPP solution 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 = int64_t;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 9;
const int INF = 0x3f3f3f3f;
const LL LLINF = 1e14;
const double pi = acos(-1.0);
char s[maxn];
LL q_pow(LL x, LL n) {
  LL ans = 1;
  x %= mod;
  while (n) {
    if (n & 1) ans = (ans * x) % mod;
    n >>= 1;
    x = (x * x) % mod;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  LL n, a, b, k;
  cin >> n >> a >> b >> k;
  cin >> s;
  LL ans = 0;
  for (int i = 0; i < k; i++) {
    if (s[i] == '+')
      ans = (ans + q_pow(a, n - i) % mod * q_pow(b, i) % mod + mod) % mod;
    else
      ans = (ans - q_pow(a, n - i) % mod * q_pow(b, i) % mod + mod) % mod;
  }
  LL t = (n + 1) / k;
  LL q = q_pow(a, k * (mod - 2) % (mod - 1)) % mod * q_pow(b, k) % mod;
  if (q == 1)
    ans = (ans * t + mod) % mod;
  else
    ans = (ans * (1 - q_pow(q, t)) % mod * q_pow(1 - q, mod - 2) % mod + mod) %
          mod;
  if (ans < 0) ans += mod;
  cout << ans;
  return 0;
}

```