#include <bits/stdc++.h>
using namespace std;
const long maxn = 1e5;
const long long mod = 1e9 + 9;
long long x, y;
long long mul(long long s, long ci) {
  long long r = 1;
  while (ci) {
    if ((ci & 1) == 1) r = r * s % mod;
    s = s * s % mod;
    ci = ci >> 1;
  }
  return r;
}
void gcd(long long a, long long b) {
  if (b == 0) {
    x = 1;
    y = 0;
  } else {
    long long r;
    gcd(b, a % b);
    r = x;
    x = y;
    y = r - a / b * y;
  }
}
long long ni(long long s) {
  gcd(mod, s);
  return (y % mod + mod) % mod;
}
int main() {
  long n, k, i;
  long long a, b, ni_a, ni_a_k, a_k, b_k, sum, c, cc, d;
  char ch;
  scanf("%ld%lld%lld%ld\n", &n, &a, &b, &k);
  c = mul(a, n);
  b_k = mul(b, k);
  a_k = mul(a, k);
  ni_a = ni(a);
  ni_a_k = ni(a_k);
  sum = 0;
  for (i = 1; i <= k; i++) {
    scanf("%c", &ch);
    if (ch == '+')
      sum = (sum + c) % mod;
    else
      sum = (sum - c + mod) % mod;
    c = c * ni_a % mod * b % mod;
  }
  d = (n + 1) / k;
  c = ni_a_k * b_k % mod;
  if (c == 1)
    printf("%lld", sum * d % mod);
  else
    printf("%lld",
           sum * ((mul(c, d) % mod - 1 + mod) % mod) % mod * ni(c - 1) % mod);
  return 0;
}

### Prompt

Please formulate 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 long maxn = 1e5;
const long long mod = 1e9 + 9;
long long x, y;
long long mul(long long s, long ci) {
  long long r = 1;
  while (ci) {
    if ((ci & 1) == 1) r = r * s % mod;
    s = s * s % mod;
    ci = ci >> 1;
  }
  return r;
}
void gcd(long long a, long long b) {
  if (b == 0) {
    x = 1;
    y = 0;
  } else {
    long long r;
    gcd(b, a % b);
    r = x;
    x = y;
    y = r - a / b * y;
  }
}
long long ni(long long s) {
  gcd(mod, s);
  return (y % mod + mod) % mod;
}
int main() {
  long n, k, i;
  long long a, b, ni_a, ni_a_k, a_k, b_k, sum, c, cc, d;
  char ch;
  scanf("%ld%lld%lld%ld\n", &n, &a, &b, &k);
  c = mul(a, n);
  b_k = mul(b, k);
  a_k = mul(a, k);
  ni_a = ni(a);
  ni_a_k = ni(a_k);
  sum = 0;
  for (i = 1; i <= k; i++) {
    scanf("%c", &ch);
    if (ch == '+')
      sum = (sum + c) % mod;
    else
      sum = (sum - c + mod) % mod;
    c = c * ni_a % mod * b % mod;
  }
  d = (n + 1) / k;
  c = ni_a_k * b_k % mod;
  if (c == 1)
    printf("%lld", sum * d % mod);
  else
    printf("%lld",
           sum * ((mul(c, d) % mod - 1 + mod) % mod) % mod * ni(c - 1) % mod);
  return 0;
}

```