#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 9;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const int N = 1e5 + 5;
int n, a, b, k;
char s[N];
long long p1[N], p2[N];
long long qmod(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
int main() {
  scanf("%d%d%d%d%s", &n, &a, &b, &k, s);
  p1[0] = p2[0] = 1;
  for (int i = 1; i < N; ++i) {
    p1[i] = p1[i - 1] * a % mod;
    p2[i] = p2[i - 1] * b % mod;
  }
  long long x0 = 0, m = (n + 1) / k - 1;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+') {
      x0 += qmod(a, n - i) * p2[i] % mod;
    } else {
      x0 -= qmod(a, n - i) * p2[i] % mod;
    }
    x0 = (x0 % mod + mod) % mod;
  }
  long long q = p2[k] * qmod(p1[k], mod - 2) % mod, ans;
  if (q == 1) {
    ans = x0 * (m + 1) % mod;
  } else {
    long long tmp =
        ((qmod(q, m + 1) - 1) % mod + mod) % mod * qmod(q - 1, mod - 2) % mod;
    ans = x0 * tmp % mod;
  }
  printf("%lld\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 mod = 1e9 + 9;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const int N = 1e5 + 5;
int n, a, b, k;
char s[N];
long long p1[N], p2[N];
long long qmod(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
int main() {
  scanf("%d%d%d%d%s", &n, &a, &b, &k, s);
  p1[0] = p2[0] = 1;
  for (int i = 1; i < N; ++i) {
    p1[i] = p1[i - 1] * a % mod;
    p2[i] = p2[i - 1] * b % mod;
  }
  long long x0 = 0, m = (n + 1) / k - 1;
  for (int i = 0; i < k; ++i) {
    if (s[i] == '+') {
      x0 += qmod(a, n - i) * p2[i] % mod;
    } else {
      x0 -= qmod(a, n - i) * p2[i] % mod;
    }
    x0 = (x0 % mod + mod) % mod;
  }
  long long q = p2[k] * qmod(p1[k], mod - 2) % mod, ans;
  if (q == 1) {
    ans = x0 * (m + 1) % mod;
  } else {
    long long tmp =
        ((qmod(q, m + 1) - 1) % mod + mod) % mod * qmod(q - 1, mod - 2) % mod;
    ans = x0 * tmp % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

```