#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 9;
long long N, A, B, K, s, ans;
char S[100002];
long long mpow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return ret;
}
long long inv(long long x) { return mpow(x, mod - 2); }
long long calc(long long x) {
  return (mod + (mpow(inv(A), x) * mpow(B, x)) % mod - 1) % mod;
}
int main() {
  scanf("%lld %lld %lld %lld", &N, &A, &B, &K);
  scanf("%s", S);
  long long invv = inv(calc(K)) % mod;
  for (int i = 0; i < K; i++) {
    s = (S[i] == '+' ? 1 : -1);
    long long x = (N - i) / K + 1;
    long long add = mpow(A, N - i) * mpow(B, i) % mod;
    if (invv > 0) {
      add = (add * calc(x * K)) % mod;
      add = (add * invv) % mod;
      add *= s;
    } else
      add = s * (add * x) % mod;
    ans = (mod + ans + add) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 long mod = 1e9 + 9;
long long N, A, B, K, s, ans;
char S[100002];
long long mpow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return ret;
}
long long inv(long long x) { return mpow(x, mod - 2); }
long long calc(long long x) {
  return (mod + (mpow(inv(A), x) * mpow(B, x)) % mod - 1) % mod;
}
int main() {
  scanf("%lld %lld %lld %lld", &N, &A, &B, &K);
  scanf("%s", S);
  long long invv = inv(calc(K)) % mod;
  for (int i = 0; i < K; i++) {
    s = (S[i] == '+' ? 1 : -1);
    long long x = (N - i) / K + 1;
    long long add = mpow(A, N - i) * mpow(B, i) % mod;
    if (invv > 0) {
      add = (add * calc(x * K)) % mod;
      add = (add * invv) % mod;
      add *= s;
    } else
      add = s * (add * x) % mod;
    ans = (mod + ans + add) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

```