#include <bits/stdc++.h>
using namespace std;
const int maxn = 110000;
const int mod = 1e9 + 9;
int pw(int x, int k) {
  int re = 1;
  for (; k; k >>= 1, x = (long long)x * x % mod)
    if (k & 1) re = (long long)re * x % mod;
  return re;
}
int inv(int x) { return pw(x, mod - 2); }
int cal(int x, int k) {
  if (x == 1) return k;
  return (1ll - pw(x, k) + mod) * inv(1ll - x + mod) % mod;
}
int n, k, a, b, ia, ib;
char str[maxn];
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", str);
  int c = (long long)pw(inv(a), k) * pw(b, k) % mod;
  int ans = 0;
  for (int i = 0; i <= n % k && i <= n; i++) {
    int j = n / k + 1;
    int x = str[i] == '+' ? 1 : -1;
    int a1 = (long long)pw(a, n - i) * pw(b, i) % mod;
    (ans += (long long)x * a1 * cal(c, j) % mod) %= mod;
  }
  for (int i = n % k + 1; i < k && i <= n; i++) {
    int j = n / k;
    int x = str[i] == '+' ? 1 : -1;
    int a1 = (long long)pw(a, n - i) * pw(b, i) % mod;
    (ans += (long long)x * a1 * cal(c, j) % mod) %= mod;
  }
  (ans += mod) %= mod;
  printf("%d\n", ans);
  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 int maxn = 110000;
const int mod = 1e9 + 9;
int pw(int x, int k) {
  int re = 1;
  for (; k; k >>= 1, x = (long long)x * x % mod)
    if (k & 1) re = (long long)re * x % mod;
  return re;
}
int inv(int x) { return pw(x, mod - 2); }
int cal(int x, int k) {
  if (x == 1) return k;
  return (1ll - pw(x, k) + mod) * inv(1ll - x + mod) % mod;
}
int n, k, a, b, ia, ib;
char str[maxn];
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", str);
  int c = (long long)pw(inv(a), k) * pw(b, k) % mod;
  int ans = 0;
  for (int i = 0; i <= n % k && i <= n; i++) {
    int j = n / k + 1;
    int x = str[i] == '+' ? 1 : -1;
    int a1 = (long long)pw(a, n - i) * pw(b, i) % mod;
    (ans += (long long)x * a1 * cal(c, j) % mod) %= mod;
  }
  for (int i = n % k + 1; i < k && i <= n; i++) {
    int j = n / k;
    int x = str[i] == '+' ? 1 : -1;
    int a1 = (long long)pw(a, n - i) * pw(b, i) % mod;
    (ans += (long long)x * a1 * cal(c, j) % mod) %= mod;
  }
  (ans += mod) %= mod;
  printf("%d\n", ans);
  return 0;
}

```