#include <bits/stdc++.h>
const int inf = (1ll << 30) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
long long a, b;
long long n, k;
char c[300100];
const int mod = 1000 * 1000 * 1000 + 9;
long long binpow(long long x, long long y) {
  x %= mod;
  x += mod;
  x %= mod;
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
long long calc(long long g, long long y) {
  if (y == 0) return 1;
  if (y % 2 == 0) return (calc(g, y - 1) + binpow(g, y)) % mod;
  return (binpow(g, (y + 1) / 2) + 1) * calc(g, y / 2) % mod;
}
void solve() {
  cin >> n >> a >> b >> k;
  scanf("%s", c);
  long long cur = binpow(a, n);
  long long ans = 0;
  b = b * binpow(a, mod - 2) % mod;
  long long g = (n + 1) / k;
  long long bb = binpow(b, k);
  for (int i = 0; i < k; i++) {
    if (c[i] == '+')
      ans = (ans + binpow(b, i) % mod) % mod;
    else
      ans = (ans - binpow(b, i) % mod) % mod;
    ans += mod;
    ans %= mod;
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  ans = (ans * 1ll * calc(bb, g - 1)) % mod;
  ans *= cur;
  ans %= mod;
  ans += mod;
  cout << ans % mod;
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  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>
const int inf = (1ll << 30) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
long long a, b;
long long n, k;
char c[300100];
const int mod = 1000 * 1000 * 1000 + 9;
long long binpow(long long x, long long y) {
  x %= mod;
  x += mod;
  x %= mod;
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
long long calc(long long g, long long y) {
  if (y == 0) return 1;
  if (y % 2 == 0) return (calc(g, y - 1) + binpow(g, y)) % mod;
  return (binpow(g, (y + 1) / 2) + 1) * calc(g, y / 2) % mod;
}
void solve() {
  cin >> n >> a >> b >> k;
  scanf("%s", c);
  long long cur = binpow(a, n);
  long long ans = 0;
  b = b * binpow(a, mod - 2) % mod;
  long long g = (n + 1) / k;
  long long bb = binpow(b, k);
  for (int i = 0; i < k; i++) {
    if (c[i] == '+')
      ans = (ans + binpow(b, i) % mod) % mod;
    else
      ans = (ans - binpow(b, i) % mod) % mod;
    ans += mod;
    ans %= mod;
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  ans = (ans * 1ll * calc(bb, g - 1)) % mod;
  ans *= cur;
  ans %= mod;
  ans += mod;
  cout << ans % mod;
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```