#include <bits/stdc++.h>
using namespace std;
const long long INF = 0x3f3f3f3f;
const double inf = 1e8 + 100;
const double eps = 1e-8;
const int maxn = 1e6;
const int N = 510;
const long long mod = 1e9 + 7;
long long x, y;
int gcd(int x, int y) {
  if (y == 0) {
    return x;
  }
  return gcd(y, x % y);
}
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = exgcd(b, a % b, x, y);
  int temp = x;
  x = y;
  y = temp - a / b * y;
  return d;
}
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = ans * a % m;
    }
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long inv(long long a) {
  exgcd(a, mod + 2, x, y);
  return (x + mod + 2) % (mod + 2);
}
void solve() {
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  char ab;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    cin >> ab;
    long long t1 = qpow(a, n - i, mod + 2);
    long long t2 = qpow(b, i, mod + 2);
    if (ab == '+') {
      ans += t1 * t2;
    } else {
      ans -= t1 * t2;
    }
    ans += (mod + 2);
    ans %= (mod + 2);
  }
  long long t = qpow(b, k, mod + 2) * qpow(inv(a), k, mod + 2) % (mod + 2);
  if (t == 1) {
    cout << (n + 1) / k * ans % (mod + 2) << endl;
  } else {
    ans = ans * (qpow(t, (n + 1) / k, mod + 2) - 1) % (mod + 2) % (mod + 2) *
          ((inv(t - 1)) % (mod + 2));
    ans = (ans) % (mod + 2);
    ans = (ans + mod + 2) % (mod + 2);
    cout << ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### 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;
const long long INF = 0x3f3f3f3f;
const double inf = 1e8 + 100;
const double eps = 1e-8;
const int maxn = 1e6;
const int N = 510;
const long long mod = 1e9 + 7;
long long x, y;
int gcd(int x, int y) {
  if (y == 0) {
    return x;
  }
  return gcd(y, x % y);
}
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = exgcd(b, a % b, x, y);
  int temp = x;
  x = y;
  y = temp - a / b * y;
  return d;
}
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) {
      ans = ans * a % m;
    }
    a = a * a % m;
    b >>= 1;
  }
  return ans;
}
long long inv(long long a) {
  exgcd(a, mod + 2, x, y);
  return (x + mod + 2) % (mod + 2);
}
void solve() {
  int n, a, b, k;
  cin >> n >> a >> b >> k;
  char ab;
  long long ans = 0;
  for (int i = 0; i < k; i++) {
    cin >> ab;
    long long t1 = qpow(a, n - i, mod + 2);
    long long t2 = qpow(b, i, mod + 2);
    if (ab == '+') {
      ans += t1 * t2;
    } else {
      ans -= t1 * t2;
    }
    ans += (mod + 2);
    ans %= (mod + 2);
  }
  long long t = qpow(b, k, mod + 2) * qpow(inv(a), k, mod + 2) % (mod + 2);
  if (t == 1) {
    cout << (n + 1) / k * ans % (mod + 2) << endl;
  } else {
    ans = ans * (qpow(t, (n + 1) / k, mod + 2) - 1) % (mod + 2) % (mod + 2) *
          ((inv(t - 1)) % (mod + 2));
    ans = (ans) % (mod + 2);
    ans = (ans + mod + 2) % (mod + 2);
    cout << ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```