#include <bits/stdc++.h>
using namespace std;
long long n, a, b, k;
long long p, f, z, x, y;
const long long mod = 1000000009;
void mult(long long a[][2], long long b[][2]) {
  int c[2][2];
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) {
      c[i][j] = 0;
      for (int k = 0; k < 2; k++)
        c[i][j] = (c[i][j] + 1LL * a[i][k] * b[k][j]) % mod;
    }
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) a[i][j] = c[i][j];
}
long long bin_pow(long long a, long long n) {
  long long r = 1;
  while (n) {
    if (n & 1) r = (1LL * r * a) % mod;
    n >>= 1;
    a = (1LL * a * a) % mod;
  }
  return r;
}
int main() {
  cin >> n >> a >> b >> k;
  p = bin_pow(a, n) % mod;
  y = bin_pow(b, k) % mod;
  x = bin_pow(a, k) % mod;
  z = y * bin_pow(x, mod - 2) % mod;
  long long A[][2] = {{1, 0}, {1, z}}, B[][2] = {{1, 0}, {0, 1}};
  long long t = (n + 1) / k;
  while (t) {
    if (t & 1) mult(B, A);
    t >>= 1;
    mult(A, A);
  }
  long long s = (1LL * p * B[1][0]) % mod;
  char c;
  f = (1LL * b * bin_pow(a, mod - 2)) % mod;
  long long ans = 0;
  for (int i = 0; i < k; ++i) {
    char c;
    cin >> c;
    if (c == '+')
      ans = (ans + s) % mod;
    else
      ans = (ans - s + mod) % mod;
    s = s * f % mod;
  }
  cout << ans;
}

### Prompt

Your task is to create a CPP solution to 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;
long long n, a, b, k;
long long p, f, z, x, y;
const long long mod = 1000000009;
void mult(long long a[][2], long long b[][2]) {
  int c[2][2];
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) {
      c[i][j] = 0;
      for (int k = 0; k < 2; k++)
        c[i][j] = (c[i][j] + 1LL * a[i][k] * b[k][j]) % mod;
    }
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) a[i][j] = c[i][j];
}
long long bin_pow(long long a, long long n) {
  long long r = 1;
  while (n) {
    if (n & 1) r = (1LL * r * a) % mod;
    n >>= 1;
    a = (1LL * a * a) % mod;
  }
  return r;
}
int main() {
  cin >> n >> a >> b >> k;
  p = bin_pow(a, n) % mod;
  y = bin_pow(b, k) % mod;
  x = bin_pow(a, k) % mod;
  z = y * bin_pow(x, mod - 2) % mod;
  long long A[][2] = {{1, 0}, {1, z}}, B[][2] = {{1, 0}, {0, 1}};
  long long t = (n + 1) / k;
  while (t) {
    if (t & 1) mult(B, A);
    t >>= 1;
    mult(A, A);
  }
  long long s = (1LL * p * B[1][0]) % mod;
  char c;
  f = (1LL * b * bin_pow(a, mod - 2)) % mod;
  long long ans = 0;
  for (int i = 0; i < k; ++i) {
    char c;
    cin >> c;
    if (c == '+')
      ans = (ans + s) % mod;
    else
      ans = (ans - s + mod) % mod;
    s = s * f % mod;
  }
  cout << ans;
}

```