#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100;
const int MOD = 1e9 + 9;
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int mul(int a, int b) { return (long long)a * b % MOD; }
int mpow(int a, int n) {
  int res = 1;
  for (; n > 0; n >>= 1) {
    if (n & 1) res = mul(res, a);
    a = mul(a, a);
  }
  return res;
}
int inv(int a) { return mpow(a, MOD - 2); }
int mdiv(int a, int b) { return mul(a, inv(b)); }
int geo(int q, int n) {
  if (q == 1) {
    return (n + 1) % MOD;
  }
  int r_n = mpow(q, n + 1);
  return mdiv((r_n - 1 + MOD) % MOD, (q - 1 + MOD) % MOD);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  int f = mdiv(b, a);
  string s;
  cin >> s;
  int ans = 0;
  int kk = (n + 1) / k;
  for (int i = 0; i < s.size(); i++) {
    int sf = s[i] == '+' ? 1 : -1;
    int q = mpow(f, i);
    int pot = geo(mpow(f, k), kk - 1);
    int foo = mul(q, pot);
    if (sf < 0) {
      foo = MOD - foo;
    }
    add(ans, foo);
  }
  ans = mul(mpow(a, n), ans);
  cout << ans << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 1e6 + 100;
const int MOD = 1e9 + 9;
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int mul(int a, int b) { return (long long)a * b % MOD; }
int mpow(int a, int n) {
  int res = 1;
  for (; n > 0; n >>= 1) {
    if (n & 1) res = mul(res, a);
    a = mul(a, a);
  }
  return res;
}
int inv(int a) { return mpow(a, MOD - 2); }
int mdiv(int a, int b) { return mul(a, inv(b)); }
int geo(int q, int n) {
  if (q == 1) {
    return (n + 1) % MOD;
  }
  int r_n = mpow(q, n + 1);
  return mdiv((r_n - 1 + MOD) % MOD, (q - 1 + MOD) % MOD);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  int f = mdiv(b, a);
  string s;
  cin >> s;
  int ans = 0;
  int kk = (n + 1) / k;
  for (int i = 0; i < s.size(); i++) {
    int sf = s[i] == '+' ? 1 : -1;
    int q = mpow(f, i);
    int pot = geo(mpow(f, k), kk - 1);
    int foo = mul(q, pot);
    if (sf < 0) {
      foo = MOD - foo;
    }
    add(ans, foo);
  }
  ans = mul(mpow(a, n), ans);
  cout << ans << '\n';
  return 0;
}

```