#include <bits/stdc++.h>
using namespace std;
int n, a, b, k;
int s[100005];
const int mod = 1e9 + 9;
int power(int a, int b) {
  int ans = 1, temp = a;
  while (b) {
    if (b & 1) ans = (1LL * ans * temp) % mod;
    temp = (1LL * temp * temp) % mod;
    b >>= 1;
  }
  return ans;
}
int exGcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int r = exGcd(b, a % b, x, y);
  int t = x;
  x = y;
  y = t - a / b * y;
  return r;
}
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  char ch = getchar();
  for (int i = 0; i < k; i++) {
    ch = getchar();
    if (ch == '+')
      s[i] = 1;
    else
      s[i] = -1;
  }
  int gg = 0;
  int aa = power(a, n), bb = 1;
  int inva, invak = power(a, k), xp, p = power(a, k);
  exGcd(a, mod, inva, xp);
  inva %= mod;
  while (inva < 0) inva += mod;
  exGcd(p, mod, p, xp);
  p %= mod;
  while (p < 0) p += mod;
  p = (1LL * p * power(b, k)) % mod;
  exGcd(invak, mod, invak, xp);
  invak %= mod;
  while (invak < 0) invak += mod;
  for (int i = 0; i < k; i++) {
    gg = (gg + 1LL * s[i] * aa * bb) % mod;
    aa = (1LL * aa * inva) % mod;
    bb = (1LL * bb * b) % mod;
  }
  int right = 0;
  gg = (gg + mod) % mod;
  int num = (n + 1) / k;
  if (p == 0) {
    printf("%d\n", gg);
    return 0;
  }
  if (p == 1) {
    printf("%d\n", ((1LL * gg * num) % mod + mod) % mod);
  } else {
    int pro = power(p, num);
    p = (1 - p) % mod;
    while (p < 0) p += mod;
    pro = (1 - pro) % mod;
    while (pro < 0) pro += mod;
    exGcd(p, mod, p, xp);
    p %= mod;
    while (p < 0) p += mod;
    gg = (1LL * gg * p) % mod;
    gg = (1LL * gg * pro) % mod;
    gg = (gg + mod) % mod;
    printf("%d\n", gg);
  }
  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;
int n, a, b, k;
int s[100005];
const int mod = 1e9 + 9;
int power(int a, int b) {
  int ans = 1, temp = a;
  while (b) {
    if (b & 1) ans = (1LL * ans * temp) % mod;
    temp = (1LL * temp * temp) % mod;
    b >>= 1;
  }
  return ans;
}
int exGcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int r = exGcd(b, a % b, x, y);
  int t = x;
  x = y;
  y = t - a / b * y;
  return r;
}
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  char ch = getchar();
  for (int i = 0; i < k; i++) {
    ch = getchar();
    if (ch == '+')
      s[i] = 1;
    else
      s[i] = -1;
  }
  int gg = 0;
  int aa = power(a, n), bb = 1;
  int inva, invak = power(a, k), xp, p = power(a, k);
  exGcd(a, mod, inva, xp);
  inva %= mod;
  while (inva < 0) inva += mod;
  exGcd(p, mod, p, xp);
  p %= mod;
  while (p < 0) p += mod;
  p = (1LL * p * power(b, k)) % mod;
  exGcd(invak, mod, invak, xp);
  invak %= mod;
  while (invak < 0) invak += mod;
  for (int i = 0; i < k; i++) {
    gg = (gg + 1LL * s[i] * aa * bb) % mod;
    aa = (1LL * aa * inva) % mod;
    bb = (1LL * bb * b) % mod;
  }
  int right = 0;
  gg = (gg + mod) % mod;
  int num = (n + 1) / k;
  if (p == 0) {
    printf("%d\n", gg);
    return 0;
  }
  if (p == 1) {
    printf("%d\n", ((1LL * gg * num) % mod + mod) % mod);
  } else {
    int pro = power(p, num);
    p = (1 - p) % mod;
    while (p < 0) p += mod;
    pro = (1 - pro) % mod;
    while (pro < 0) pro += mod;
    exGcd(p, mod, p, xp);
    p %= mod;
    while (p < 0) p += mod;
    gg = (1LL * gg * p) % mod;
    gg = (1LL * gg * pro) % mod;
    gg = (gg + mod) % mod;
    printf("%d\n", gg);
  }
  return 0;
}

```