#include <bits/stdc++.h>
const long long MOD = 1000000009;
long long logpow(long long a, long long n) {
  long long p = 1;
  while (n) {
    if (n % 2) p = (p * a) % MOD;
    a = (a * a) % MOD;
    n /= 2;
  }
  return p;
}
char v[1 + 100000];
int main() {
  FILE *fi, *fo;
  fi = stdin;
  fo = stdout;
  long long n, a, b, k;
  fscanf(fi, "%lld%lld%lld%lld", &n, &a, &b, &k);
  fgetc(fi);
  for (int i = 0; i < k; i++) {
    v[i] = fgetc(fi);
    v[i] = (v[i] == '+' ? 1 : -1);
  }
  long long ans = 0;
  long long A = logpow(a, n), B = 1;
  long long invA = logpow(a, MOD - 2), invB = logpow(b, MOD - 2);
  for (int i = 0; i < k; i++) {
    ans = (ans + v[i] * A * B + MOD * MOD) % MOD;
    A = (A * invA) % MOD;
    B = (B * b) % MOD;
  }
  long long x = logpow((b * invA) % MOD, k);
  if (x != 1) {
    long long tmp = (logpow(x, (n + 1) / k) + MOD - 1) % MOD;
    long long TMP = logpow((x + MOD - 1) % MOD, MOD - 2);
    tmp = (tmp * TMP) % MOD;
    fprintf(fo, "%lld", (ans * tmp) % MOD);
  } else {
    fprintf(fo, "%lld", (ans * (n + 1) / k) % MOD);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 long long MOD = 1000000009;
long long logpow(long long a, long long n) {
  long long p = 1;
  while (n) {
    if (n % 2) p = (p * a) % MOD;
    a = (a * a) % MOD;
    n /= 2;
  }
  return p;
}
char v[1 + 100000];
int main() {
  FILE *fi, *fo;
  fi = stdin;
  fo = stdout;
  long long n, a, b, k;
  fscanf(fi, "%lld%lld%lld%lld", &n, &a, &b, &k);
  fgetc(fi);
  for (int i = 0; i < k; i++) {
    v[i] = fgetc(fi);
    v[i] = (v[i] == '+' ? 1 : -1);
  }
  long long ans = 0;
  long long A = logpow(a, n), B = 1;
  long long invA = logpow(a, MOD - 2), invB = logpow(b, MOD - 2);
  for (int i = 0; i < k; i++) {
    ans = (ans + v[i] * A * B + MOD * MOD) % MOD;
    A = (A * invA) % MOD;
    B = (B * b) % MOD;
  }
  long long x = logpow((b * invA) % MOD, k);
  if (x != 1) {
    long long tmp = (logpow(x, (n + 1) / k) + MOD - 1) % MOD;
    long long TMP = logpow((x + MOD - 1) % MOD, MOD - 2);
    tmp = (tmp * TMP) % MOD;
    fprintf(fo, "%lld", (ans * tmp) % MOD);
  } else {
    fprintf(fo, "%lld", (ans * (n + 1) / k) % MOD);
  }
  return 0;
}

```