#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
const long long mo = 1000000009;
long long N, A, B, K;
char S[maxn];
long long qkpow(long long x, long long y) {
  long long re = 1, t = x;
  while (y) {
    if (y & 1) re = 1ll * re * t % mo;
    t = 1ll * t * t % mo;
    y >>= 1;
  }
  return re;
}
long long Inv(long long x) { return qkpow(x, mo - 2); }
int main() {
  scanf("%lld%lld%lld%lld%s", &N, &A, &B, &K, S);
  long long sum = 0, pa = qkpow(A, N), pb = 1, inva = Inv(A);
  for (int i = 0; i < K; i++) {
    long long sign = S[i] == '+' ? 1 : -1;
    sum = ((sum + sign * pa * pb % mo) % mo + mo) % mo;
    pa = 1ll * pa * inva % mo;
    pb = 1ll * pb * B % mo;
  }
  long long val = 1ll * qkpow(inva, K) * qkpow(B, K) % mo;
  long long ans = 0;
  if (val != 1)
    ans = 1ll * (1 - qkpow(val, (N + 1) / K) + mo) % mo *
          Inv((1 - val + mo) % mo) % mo * sum % mo;
  else
    ans = 1ll * sum * (N + 1) / K % mo;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 100005;
const long long mo = 1000000009;
long long N, A, B, K;
char S[maxn];
long long qkpow(long long x, long long y) {
  long long re = 1, t = x;
  while (y) {
    if (y & 1) re = 1ll * re * t % mo;
    t = 1ll * t * t % mo;
    y >>= 1;
  }
  return re;
}
long long Inv(long long x) { return qkpow(x, mo - 2); }
int main() {
  scanf("%lld%lld%lld%lld%s", &N, &A, &B, &K, S);
  long long sum = 0, pa = qkpow(A, N), pb = 1, inva = Inv(A);
  for (int i = 0; i < K; i++) {
    long long sign = S[i] == '+' ? 1 : -1;
    sum = ((sum + sign * pa * pb % mo) % mo + mo) % mo;
    pa = 1ll * pa * inva % mo;
    pb = 1ll * pb * B % mo;
  }
  long long val = 1ll * qkpow(inva, K) * qkpow(B, K) % mo;
  long long ans = 0;
  if (val != 1)
    ans = 1ll * (1 - qkpow(val, (N + 1) / K) + mo) % mo *
          Inv((1 - val + mo) % mo) % mo * sum % mo;
  else
    ans = 1ll * sum * (N + 1) / K % mo;
  printf("%lld\n", ans);
  return 0;
}

```