#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y, long long int m) {
  long long int a = 1;
  x %= m;
  while (y > 0) {
    if (y & 1) {
      a *= x;
      a %= m;
    }
    y /= 2;
    x *= x;
    x %= m;
  }
  return a;
}
long long int modInverse(long long int b, long long int m) {
  return power(b, m - 2, m);
}
long long int modDivide(long long int a, long long int b, long long int m) {
  a %= m;
  long long int inv = modInverse(b, m);
  a *= inv;
  a %= m;
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, a, b, k;
  long long int mod = 1000000009;
  cin >> n >> a >> b >> k;
  vector<long long int> v;
  long long int x = power(a, n, mod);
  long long int s = 0;
  for (long long int i = 0; i < k; i++) {
    char c;
    cin >> c;
    if (c == '+') {
      v.push_back(1);
    } else {
      v.push_back(-1);
    }
    long long int xx = x * v[i];
    s += xx;
    s += mod;
    s %= mod;
    x *= b;
    x %= mod;
    x = modDivide(x, a, mod);
  }
  long long int f = (n + 1) / k;
  long long int k1 = k * f;
  x = power(b, k1, mod);
  long long int y = power(a, k1, mod);
  long long int r = power(a, k, mod);
  long long int r1 = power(b, k, mod);
  if (r1 == r) {
    s *= f;
    s %= mod;
    cout << s << endl;
    return 0;
  }
  x += mod;
  x -= y;
  x %= mod;
  s *= x;
  s %= mod;
  s *= r;
  s %= mod;
  s = modDivide(s, y, mod);
  r1 += mod;
  r1 -= r;
  r1 %= mod;
  s = modDivide(s, r1, mod);
  cout << s << endl;
}

### Prompt

Generate 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;
long long int power(long long int x, long long int y, long long int m) {
  long long int a = 1;
  x %= m;
  while (y > 0) {
    if (y & 1) {
      a *= x;
      a %= m;
    }
    y /= 2;
    x *= x;
    x %= m;
  }
  return a;
}
long long int modInverse(long long int b, long long int m) {
  return power(b, m - 2, m);
}
long long int modDivide(long long int a, long long int b, long long int m) {
  a %= m;
  long long int inv = modInverse(b, m);
  a *= inv;
  a %= m;
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, a, b, k;
  long long int mod = 1000000009;
  cin >> n >> a >> b >> k;
  vector<long long int> v;
  long long int x = power(a, n, mod);
  long long int s = 0;
  for (long long int i = 0; i < k; i++) {
    char c;
    cin >> c;
    if (c == '+') {
      v.push_back(1);
    } else {
      v.push_back(-1);
    }
    long long int xx = x * v[i];
    s += xx;
    s += mod;
    s %= mod;
    x *= b;
    x %= mod;
    x = modDivide(x, a, mod);
  }
  long long int f = (n + 1) / k;
  long long int k1 = k * f;
  x = power(b, k1, mod);
  long long int y = power(a, k1, mod);
  long long int r = power(a, k, mod);
  long long int r1 = power(b, k, mod);
  if (r1 == r) {
    s *= f;
    s %= mod;
    cout << s << endl;
    return 0;
  }
  x += mod;
  x -= y;
  x %= mod;
  s *= x;
  s %= mod;
  s *= r;
  s %= mod;
  s = modDivide(s, y, mod);
  r1 += mod;
  r1 -= r;
  r1 %= mod;
  s = modDivide(s, r1, mod);
  cout << s << endl;
}

```