#include <bits/stdc++.h>
using namespace std;
const long long N = 200010, Mod = 1e9 + 9;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long n, A, B, K, gb, res;
char inp[N];
inline long long ksc(long long x, long long p) {
  long long res = 1;
  for (; p; p >>= 1, x = x * x % Mod)
    if (p & 1) res = res * x % Mod;
  return res;
}
signed main() {
  n = read(), A = read(), B = read(), K = read();
  gb = ksc(B * ksc(A, Mod - 2) % Mod, K) % Mod;
  scanf("%s", inp);
  for (register long long i = 0; i < K; i++) {
    long long up = (inp[i] == '+') ? (1) : (-1);
    long long fir = ksc(A, n - i) * ksc(B, i) % Mod;
    long long m = (n - i) / K + 1;
    if (gb == 1)
      res = (res + up * m * fir % Mod + Mod) % Mod;
    else
      res = (res +
             up * fir % Mod * (1 - ksc(gb, m) + Mod) % Mod *
                 ksc((1 - gb + Mod) % Mod, Mod - 2) % Mod +
             Mod) %
            Mod;
  }
  printf("%lld\n", res);
  return 0;
}

### Prompt

Develop a solution in CPP 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 long long N = 200010, Mod = 1e9 + 9;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long n, A, B, K, gb, res;
char inp[N];
inline long long ksc(long long x, long long p) {
  long long res = 1;
  for (; p; p >>= 1, x = x * x % Mod)
    if (p & 1) res = res * x % Mod;
  return res;
}
signed main() {
  n = read(), A = read(), B = read(), K = read();
  gb = ksc(B * ksc(A, Mod - 2) % Mod, K) % Mod;
  scanf("%s", inp);
  for (register long long i = 0; i < K; i++) {
    long long up = (inp[i] == '+') ? (1) : (-1);
    long long fir = ksc(A, n - i) * ksc(B, i) % Mod;
    long long m = (n - i) / K + 1;
    if (gb == 1)
      res = (res + up * m * fir % Mod + Mod) % Mod;
    else
      res = (res +
             up * fir % Mod * (1 - ksc(gb, m) + Mod) % Mod *
                 ksc((1 - gb + Mod) % Mod, Mod - 2) % Mod +
             Mod) %
            Mod;
  }
  printf("%lld\n", res);
  return 0;
}

```