#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long m = 1e9 + 9;
long long add(long long x, long long y) {
  x += y;
  x %= m;
  while (x < 0) x += m;
  return x;
}
long long mul(long long x, long long y) {
  long long ans = (x * 1ll * y) % m;
  while (ans < 0) ans += m;
  return ans;
}
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
long long inv(long long x) { return binpow(x, m - 2); }
long long divide(long long x, long long y) { return mul(x, inv(y)); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n = 0, a = 0, b = 0, k = 0;
  cin >> n >> a >> b >> k;
  n++;
  string s;
  cin >> s;
  long long cur = 1, ans = 0;
  long long c = mul(b, inv(a));
  for (int i = 0; i < k; i++) {
    ans += cur * (s[i] == '+' ? 1 : -1);
    cur *= c;
    ans %= m;
    cur %= m;
  }
  if (cur == 1)
    ans = ans * n / k;
  else
    ans = ans * (binpow(cur, n / k) - 1) % m * inv(cur - 1) % m;
  ans %= m;
  ans = ans * binpow(a, n - 1) % m;
  while (ans < m) ans += m;
  cout << ans % m;
  return 0;
}

### Prompt

Please create a solution in CPP 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>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long m = 1e9 + 9;
long long add(long long x, long long y) {
  x += y;
  x %= m;
  while (x < 0) x += m;
  return x;
}
long long mul(long long x, long long y) {
  long long ans = (x * 1ll * y) % m;
  while (ans < 0) ans += m;
  return ans;
}
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
long long inv(long long x) { return binpow(x, m - 2); }
long long divide(long long x, long long y) { return mul(x, inv(y)); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n = 0, a = 0, b = 0, k = 0;
  cin >> n >> a >> b >> k;
  n++;
  string s;
  cin >> s;
  long long cur = 1, ans = 0;
  long long c = mul(b, inv(a));
  for (int i = 0; i < k; i++) {
    ans += cur * (s[i] == '+' ? 1 : -1);
    cur *= c;
    ans %= m;
    cur %= m;
  }
  if (cur == 1)
    ans = ans * n / k;
  else
    ans = ans * (binpow(cur, n / k) - 1) % m * inv(cur - 1) % m;
  ans %= m;
  ans = ans * binpow(a, n - 1) % m;
  while (ans < m) ans += m;
  cout << ans % m;
  return 0;
}

```