#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000009;
const int maxn = 100000 + 100;
long long n, a, b, k;
char str[maxn];
long long fast_pow(long long res, long long n) {
  long long ans;
  for (ans = 1; n != 0; n >>= 1) {
    if (n % 2 == 1) {
      ans = (ans * res) % MOD;
    }
    res = (res * res) % MOD;
  }
  return ans;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, y, x);
  y -= x * (a / b);
}
long long inv(long long x) {
  long long xx, yy;
  exgcd(x, MOD, xx, yy);
  xx = ((xx % MOD) + MOD) % MOD;
  return xx;
}
int main() {
  while (scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k) != EOF) {
    if (k >= n + 1) {
      k = n + 1;
    }
    scanf("%s", str);
    long long tmp = 0;
    long long inva = inv(a);
    long long invb = inv(b);
    long long aa = fast_pow(a, n + 1);
    long long bb = invb;
    for (long long i = 0; i < k; ++i) {
      aa = (aa * inva) % MOD;
      bb = (bb * b) % MOD;
      if (str[i] == '+') {
        tmp += (aa * bb) % MOD;
      } else {
        tmp -= (aa * bb) % MOD;
      }
      tmp = ((tmp % MOD) + MOD) % MOD;
    }
    long long q = fast_pow((inva * b) % MOD, k);
    long long nn = (n + 1) / k;
    if (q == 1) {
      tmp = (nn * tmp) % MOD;
    } else {
      long long fenzi = (((1 - fast_pow(q, nn)) % MOD) + MOD) % MOD;
      long long fenmu = ((1 - q) % MOD + MOD) % MOD;
      fenzi = (fenzi * inv(fenmu)) % MOD;
      tmp = (tmp * fenzi) % MOD;
    }
    printf("%I64d\n", tmp);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 MOD = 1000000009;
const int maxn = 100000 + 100;
long long n, a, b, k;
char str[maxn];
long long fast_pow(long long res, long long n) {
  long long ans;
  for (ans = 1; n != 0; n >>= 1) {
    if (n % 2 == 1) {
      ans = (ans * res) % MOD;
    }
    res = (res * res) % MOD;
  }
  return ans;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, y, x);
  y -= x * (a / b);
}
long long inv(long long x) {
  long long xx, yy;
  exgcd(x, MOD, xx, yy);
  xx = ((xx % MOD) + MOD) % MOD;
  return xx;
}
int main() {
  while (scanf("%I64d%I64d%I64d%I64d", &n, &a, &b, &k) != EOF) {
    if (k >= n + 1) {
      k = n + 1;
    }
    scanf("%s", str);
    long long tmp = 0;
    long long inva = inv(a);
    long long invb = inv(b);
    long long aa = fast_pow(a, n + 1);
    long long bb = invb;
    for (long long i = 0; i < k; ++i) {
      aa = (aa * inva) % MOD;
      bb = (bb * b) % MOD;
      if (str[i] == '+') {
        tmp += (aa * bb) % MOD;
      } else {
        tmp -= (aa * bb) % MOD;
      }
      tmp = ((tmp % MOD) + MOD) % MOD;
    }
    long long q = fast_pow((inva * b) % MOD, k);
    long long nn = (n + 1) / k;
    if (q == 1) {
      tmp = (nn * tmp) % MOD;
    } else {
      long long fenzi = (((1 - fast_pow(q, nn)) % MOD) + MOD) % MOD;
      long long fenmu = ((1 - q) % MOD + MOD) % MOD;
      fenzi = (fenzi * inv(fenmu)) % MOD;
      tmp = (tmp * fenzi) % MOD;
    }
    printf("%I64d\n", tmp);
  }
  return 0;
}

```