#include <bits/stdc++.h>
using namespace std;
char s[100010];
void extgcd(long long a, long long b, long long d, long long &x, long long &y) {
  if (b == 0) {
    d = a;
    x = 1;
    y = 0;
  } else {
    extgcd(b, a % b, d, y, x);
    y = y - x * (a / b);
  }
}
long long inverse(long long a, long long p) {
  long long d, x, y;
  extgcd(a, p, d, x, y);
  return (x % p + p) % p;
}
long long PowerMod(long long a, long long b, long long c) {
  long long ans = 1;
  long long k;
  k = a;
  k = k % c;
  while (b > 0) {
    if (b & 1) ans = (ans * k) % c;
    b >>= 1;
    k = (k * k) % c;
  }
  return ans;
}
int main() {
  long long n, a, b, k;
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long xgb = (b * inverse(a, 1000000009)) % 1000000009;
  long long dgb = PowerMod(xgb, k, 1000000009);
  long long ffir = PowerMod(a, n, 1000000009);
  long long temp = ffir;
  long long fir = 0;
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+')
      fir = (fir + temp) % 1000000009;
    else if (s[i] == '-')
      fir = (fir - temp + 1000000009) % 1000000009;
    temp = (temp * xgb) % 1000000009;
  }
  long long cd = (n + 1) / k;
  long long ls = (n + 1) / k * k;
  long long ans = 0;
  if (PowerMod(dgb, cd, 1000000009) == 1)
    ans = (ans + (fir * cd) % 1000000009) % 1000000009;
  else
    ans = (ans + ((fir * (1 - PowerMod(dgb, cd, 1000000009) + 1000000009) %
                   1000000009) %
                  1000000009 *
                  inverse((1 - dgb + 1000000009) % 1000000009, 1000000009)) %
                     1000000009) %
          1000000009;
  temp = (ffir * (PowerMod(xgb, ls, 1000000009)));
  for (long long i = ls; i <= n; i++) {
    if (s[i - ls] == '+')
      ans = (ans + temp) % 1000000009;
    else if (s[i - ls] == '-')
      ans = (ans - temp) % 1000000009;
    temp = (temp * xgb) % 1000000009;
  }
  printf("%lld\n", ans);
  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;
char s[100010];
void extgcd(long long a, long long b, long long d, long long &x, long long &y) {
  if (b == 0) {
    d = a;
    x = 1;
    y = 0;
  } else {
    extgcd(b, a % b, d, y, x);
    y = y - x * (a / b);
  }
}
long long inverse(long long a, long long p) {
  long long d, x, y;
  extgcd(a, p, d, x, y);
  return (x % p + p) % p;
}
long long PowerMod(long long a, long long b, long long c) {
  long long ans = 1;
  long long k;
  k = a;
  k = k % c;
  while (b > 0) {
    if (b & 1) ans = (ans * k) % c;
    b >>= 1;
    k = (k * k) % c;
  }
  return ans;
}
int main() {
  long long n, a, b, k;
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long xgb = (b * inverse(a, 1000000009)) % 1000000009;
  long long dgb = PowerMod(xgb, k, 1000000009);
  long long ffir = PowerMod(a, n, 1000000009);
  long long temp = ffir;
  long long fir = 0;
  for (long long i = 0; i < k; i++) {
    if (s[i] == '+')
      fir = (fir + temp) % 1000000009;
    else if (s[i] == '-')
      fir = (fir - temp + 1000000009) % 1000000009;
    temp = (temp * xgb) % 1000000009;
  }
  long long cd = (n + 1) / k;
  long long ls = (n + 1) / k * k;
  long long ans = 0;
  if (PowerMod(dgb, cd, 1000000009) == 1)
    ans = (ans + (fir * cd) % 1000000009) % 1000000009;
  else
    ans = (ans + ((fir * (1 - PowerMod(dgb, cd, 1000000009) + 1000000009) %
                   1000000009) %
                  1000000009 *
                  inverse((1 - dgb + 1000000009) % 1000000009, 1000000009)) %
                     1000000009) %
          1000000009;
  temp = (ffir * (PowerMod(xgb, ls, 1000000009)));
  for (long long i = ls; i <= n; i++) {
    if (s[i - ls] == '+')
      ans = (ans + temp) % 1000000009;
    else if (s[i - ls] == '-')
      ans = (ans - temp) % 1000000009;
    temp = (temp * xgb) % 1000000009;
  }
  printf("%lld\n", ans);
  return 0;
}

```