#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 2;
const int mo = 1e9 + 9;
long long n, a, b, k;
char s[maxn];
long long fstpw(long long x, long long t) {
  long long ret = 1;
  while (t) {
    if (t & 1) {
      ret = (ret * x) % mo;
    }
    x = (x * x) % mo;
    t >>= 1;
  }
  return ret;
}
long long exgcd(long long aa, long long bb, long long &x, long long &y) {
  long long ret, tmp;
  if (!bb) {
    x = 1;
    y = 0;
    return a;
  }
  ret = exgcd(bb, aa % bb, x, y);
  tmp = x;
  x = y;
  y = tmp - aa / bb * y;
  return ret;
}
long long mult(long long aa, long long bb) {
  long long x, y;
  exgcd(aa, bb, x, y);
  return (x % mo + mo) % mo;
}
int main() {
  scanf("%lld%lld%lld%lld%s", &n, &a, &b, &k, s);
  long long prt = 0;
  for (int i = 0; i < k; ++i) {
    long long one = (s[i] == '+') ? 1 : -1;
    prt = (prt + one * fstpw(a, n - i) % mo * fstpw(b, i) % mo) % mo;
  }
  long long m = (n + 1) / k, q = fstpw(b, k) * mult(fstpw(a, k), mo) % mo, ans;
  if (q == 1) {
    ans = prt * m % mo;
  } else {
    ans = prt * (fstpw(q, m) - 1) % mo * mult(q - 1, mo) % mo;
  }
  printf("%lld\n", (ans + mo) % mo);
  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>
using namespace std;
const int maxn = 1e5 + 2;
const int mo = 1e9 + 9;
long long n, a, b, k;
char s[maxn];
long long fstpw(long long x, long long t) {
  long long ret = 1;
  while (t) {
    if (t & 1) {
      ret = (ret * x) % mo;
    }
    x = (x * x) % mo;
    t >>= 1;
  }
  return ret;
}
long long exgcd(long long aa, long long bb, long long &x, long long &y) {
  long long ret, tmp;
  if (!bb) {
    x = 1;
    y = 0;
    return a;
  }
  ret = exgcd(bb, aa % bb, x, y);
  tmp = x;
  x = y;
  y = tmp - aa / bb * y;
  return ret;
}
long long mult(long long aa, long long bb) {
  long long x, y;
  exgcd(aa, bb, x, y);
  return (x % mo + mo) % mo;
}
int main() {
  scanf("%lld%lld%lld%lld%s", &n, &a, &b, &k, s);
  long long prt = 0;
  for (int i = 0; i < k; ++i) {
    long long one = (s[i] == '+') ? 1 : -1;
    prt = (prt + one * fstpw(a, n - i) % mo * fstpw(b, i) % mo) % mo;
  }
  long long m = (n + 1) / k, q = fstpw(b, k) * mult(fstpw(a, k), mo) % mo, ans;
  if (q == 1) {
    ans = prt * m % mo;
  } else {
    ans = prt * (fstpw(q, m) - 1) % mo * mult(q - 1, mo) % mo;
  }
  printf("%lld\n", (ans + mo) % mo);
  return 0;
}

```