#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 9;
inline long long qp(long long a, long long n) {
  if (n == 0) return 1ll;
  long long ans = qp(a, n >> 1);
  ans *= ans;
  ans %= MOD;
  if (n & 1) ans *= a;
  ans %= MOD;
  return ans;
}
long long inv(long long n) { return qp(n, MOD - 2ll); }
char ss[110100];
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", ss);
  long long sml = 0ll;
  for (int i = 0; i < k; i++) {
    if (ss[i] == '+') {
      long long temp = 1ll * qp(a, n - i);
      temp %= MOD;
      temp *= qp(b, i);
      temp %= MOD;
      sml += temp;
      sml %= MOD;
    } else {
      long long temp = 1ll * qp(a, n - i);
      temp %= MOD;
      temp *= qp(b, i);
      temp %= MOD;
      temp = MOD - temp;
      sml += temp;
      sml %= MOD;
    }
  }
  long long aa = qp(a, k);
  long long bb = qp(b, k);
  long long ba = bb * inv(aa);
  ba %= MOD;
  if (ba == 1ll) {
    long long nn = (n + 1) / k;
    sml *= nn;
    sml %= MOD;
    printf("%lld\n", sml);
    return 0;
  }
  long long nn = (n + 1) / k;
  long long ans = sml;
  long long y = MOD + 1ll - qp(ba, nn);
  y %= MOD;
  y += MOD;
  y %= MOD;
  long long x = MOD + 1ll - ba;
  x %= MOD;
  x += MOD;
  x %= MOD;
  while (x == 0ll)
    ;
  ans *= y;
  ans %= MOD;
  ans *= inv(x);
  ans %= MOD;
  ans += MOD;
  ans %= MOD;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your task is to create 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;
inline long long qp(long long a, long long n) {
  if (n == 0) return 1ll;
  long long ans = qp(a, n >> 1);
  ans *= ans;
  ans %= MOD;
  if (n & 1) ans *= a;
  ans %= MOD;
  return ans;
}
long long inv(long long n) { return qp(n, MOD - 2ll); }
char ss[110100];
int main() {
  int n, a, b, k;
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", ss);
  long long sml = 0ll;
  for (int i = 0; i < k; i++) {
    if (ss[i] == '+') {
      long long temp = 1ll * qp(a, n - i);
      temp %= MOD;
      temp *= qp(b, i);
      temp %= MOD;
      sml += temp;
      sml %= MOD;
    } else {
      long long temp = 1ll * qp(a, n - i);
      temp %= MOD;
      temp *= qp(b, i);
      temp %= MOD;
      temp = MOD - temp;
      sml += temp;
      sml %= MOD;
    }
  }
  long long aa = qp(a, k);
  long long bb = qp(b, k);
  long long ba = bb * inv(aa);
  ba %= MOD;
  if (ba == 1ll) {
    long long nn = (n + 1) / k;
    sml *= nn;
    sml %= MOD;
    printf("%lld\n", sml);
    return 0;
  }
  long long nn = (n + 1) / k;
  long long ans = sml;
  long long y = MOD + 1ll - qp(ba, nn);
  y %= MOD;
  y += MOD;
  y %= MOD;
  long long x = MOD + 1ll - ba;
  x %= MOD;
  x += MOD;
  x %= MOD;
  while (x == 0ll)
    ;
  ans *= y;
  ans %= MOD;
  ans *= inv(x);
  ans %= MOD;
  ans += MOD;
  ans %= MOD;
  printf("%lld\n", ans);
  return 0;
}

```