#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
long long n, a, b;
int k, sig[maxn];
const long long mod = 1e9 + 9;
char s[maxn];
long long Sub(long long x, long long y) {
  long long t = (x - y) % mod;
  if (t < 0) t += mod;
  return t;
}
long long Mul(long long x, long long y) { return x * y % mod; }
long long Add(long long x, long long y) {
  long long t = (x + y) % mod;
  if (t < 0) t += mod;
  return t;
}
long long qpow(long long x, long long a) {
  long long ret = 1;
  while (a) {
    if (a & 1) ret = Mul(ret, x);
    x = Mul(x, x);
    a >>= 1;
  }
  return ret;
}
long long inv(long long x) { return qpow(x, mod - 2); }
long long Div(long long x, long long y) { return Mul(x, qpow(y, mod - 2)); }
long long getSum(long long a1, long long q, long long n) {
  if (q == 1LL) return Mul(a1, n);
  return Div(Mul(a1, Sub(1LL, qpow(q, n))), Sub(1, q));
}
int main() {
  scanf("%lld%lld%lld%d", &n, &a, &b, &k);
  scanf("%s", s);
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+')
      sig[i] = 1;
    else
      sig[i] = -1;
  }
  long long res = 0, ans = 0, invA = inv(qpow(a, k)), B = qpow(b, k);
  for (int i = 0; i < k; i++)
    res = Add(res, (long long)sig[i] * Mul(qpow(a, n - i), qpow(b, i)));
  printf("%lld\n", getSum(res, Mul(B, invA), ((long long)n + 1) / k));
}

### 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 + 10;
long long n, a, b;
int k, sig[maxn];
const long long mod = 1e9 + 9;
char s[maxn];
long long Sub(long long x, long long y) {
  long long t = (x - y) % mod;
  if (t < 0) t += mod;
  return t;
}
long long Mul(long long x, long long y) { return x * y % mod; }
long long Add(long long x, long long y) {
  long long t = (x + y) % mod;
  if (t < 0) t += mod;
  return t;
}
long long qpow(long long x, long long a) {
  long long ret = 1;
  while (a) {
    if (a & 1) ret = Mul(ret, x);
    x = Mul(x, x);
    a >>= 1;
  }
  return ret;
}
long long inv(long long x) { return qpow(x, mod - 2); }
long long Div(long long x, long long y) { return Mul(x, qpow(y, mod - 2)); }
long long getSum(long long a1, long long q, long long n) {
  if (q == 1LL) return Mul(a1, n);
  return Div(Mul(a1, Sub(1LL, qpow(q, n))), Sub(1, q));
}
int main() {
  scanf("%lld%lld%lld%d", &n, &a, &b, &k);
  scanf("%s", s);
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+')
      sig[i] = 1;
    else
      sig[i] = -1;
  }
  long long res = 0, ans = 0, invA = inv(qpow(a, k)), B = qpow(b, k);
  for (int i = 0; i < k; i++)
    res = Add(res, (long long)sig[i] * Mul(qpow(a, n - i), qpow(b, i)));
  printf("%lld\n", getSum(res, Mul(B, invA), ((long long)n + 1) / k));
}

```