#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 9;
char s[MAXN];
int n, a, b, k;
inline int add(const int &a, const int &b) {
  if (a + b >= MOD) return a + b - MOD;
  if (a + b < 0) return a + b + MOD;
  return a + b;
}
inline int mul(const int &a, const int &b) { return (long long)a * b % MOD; }
int mypow(int b, int e) {
  int ret = 1;
  while (e) {
    if (e & 1) {
      ret = mul(ret, b);
      --e;
    }
    b = mul(b, b);
    e /= 2;
  }
  return ret;
}
int inv(const int &x) { return mypow(x, MOD - 2); }
int geometric(int a0, int r, int n) {
  if (r == 1) return mul(a0, n);
  return mul(a0, mul(add(1, -mypow(r, n)), inv(add(1, -r))));
}
int main(void) {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", s);
  int sol = 0;
  for (int i = 0; i < k; ++i) {
    int a0 = mul(mypow(a, n - i), mypow(b, i));
    int r = mul(mypow(b, k), inv(mypow(a, k)));
    if (s[i] == '+')
      sol = add(sol, geometric(a0, r, (n + 1) / k));
    else
      sol = add(sol, -geometric(a0, r, (n + 1) / k));
  }
  printf("%d\n", sol);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 int MAXN = 1e5 + 10;
const int MOD = 1e9 + 9;
char s[MAXN];
int n, a, b, k;
inline int add(const int &a, const int &b) {
  if (a + b >= MOD) return a + b - MOD;
  if (a + b < 0) return a + b + MOD;
  return a + b;
}
inline int mul(const int &a, const int &b) { return (long long)a * b % MOD; }
int mypow(int b, int e) {
  int ret = 1;
  while (e) {
    if (e & 1) {
      ret = mul(ret, b);
      --e;
    }
    b = mul(b, b);
    e /= 2;
  }
  return ret;
}
int inv(const int &x) { return mypow(x, MOD - 2); }
int geometric(int a0, int r, int n) {
  if (r == 1) return mul(a0, n);
  return mul(a0, mul(add(1, -mypow(r, n)), inv(add(1, -r))));
}
int main(void) {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", s);
  int sol = 0;
  for (int i = 0; i < k; ++i) {
    int a0 = mul(mypow(a, n - i), mypow(b, i));
    int r = mul(mypow(b, k), inv(mypow(a, k)));
    if (s[i] == '+')
      sol = add(sol, geometric(a0, r, (n + 1) / k));
    else
      sol = add(sol, -geometric(a0, r, (n + 1) / k));
  }
  printf("%d\n", sol);
  return 0;
}

```