#include <bits/stdc++.h>
using namespace std;
long long int bigMod(long long int n, long long int r, long long int mod) {
  if (!n) return 0;
  if (!r) return 1;
  if (r % 2) return (bigMod(n, r - 1, mod) * n) % mod;
  long long int ans = bigMod(n, r / 2, mod);
  return (ans * ans) % mod;
}
long long int modInverse(long long int n, long long int k) {
  return bigMod(n, k - 2, k);
}
long long int n, a, b, k, sn[200005];
string seq;
long long int X, Y;
int main() {
  int T = 1;
  while (T--) {
    scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
    cin >> seq;
    k = seq.size();
    for (long long int i = 0; i < k; i++) {
      if (seq[i] == '+')
        sn[i] = +1;
      else
        sn[i] = -1;
    }
    X = Y = 0;
    long long int ans = 0;
    long long int keep = 0;
    ;
    ans = 0;
    for (long long int i = 0; i < k; i++) {
      X = (X + sn[i] * (bigMod(a, n - i, 1000000009LL) *
                        bigMod(b, i, 1000000009LL))) %
          1000000009LL;
    }
    X = (X + 1000000009LL) % 1000000009LL;
    Y = bigMod((b * modInverse(a, 1000000009LL)) % 1000000009LL, k,
               1000000009LL);
    long long int up = (n + 1) / k;
    if (Y == 1) {
      cout << (X * up) % 1000000009LL << endl;
      return 0;
    }
    long long int temp =
        (((X * (bigMod(Y, up, 1000000009LL) - 1)) % 1000000009LL) *
             (max(1LL, modInverse((Y - 1 + 1000000009LL) % 1000000009LL,
                                  1000000009LL)) %
              1000000009LL) +
         1000000009LL) %
        1000000009LL;
    ans = temp;
    cout << ans << endl;
  }
}

### Prompt

Your challenge is to write 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 int bigMod(long long int n, long long int r, long long int mod) {
  if (!n) return 0;
  if (!r) return 1;
  if (r % 2) return (bigMod(n, r - 1, mod) * n) % mod;
  long long int ans = bigMod(n, r / 2, mod);
  return (ans * ans) % mod;
}
long long int modInverse(long long int n, long long int k) {
  return bigMod(n, k - 2, k);
}
long long int n, a, b, k, sn[200005];
string seq;
long long int X, Y;
int main() {
  int T = 1;
  while (T--) {
    scanf("%lld %lld %lld %lld", &n, &a, &b, &k);
    cin >> seq;
    k = seq.size();
    for (long long int i = 0; i < k; i++) {
      if (seq[i] == '+')
        sn[i] = +1;
      else
        sn[i] = -1;
    }
    X = Y = 0;
    long long int ans = 0;
    long long int keep = 0;
    ;
    ans = 0;
    for (long long int i = 0; i < k; i++) {
      X = (X + sn[i] * (bigMod(a, n - i, 1000000009LL) *
                        bigMod(b, i, 1000000009LL))) %
          1000000009LL;
    }
    X = (X + 1000000009LL) % 1000000009LL;
    Y = bigMod((b * modInverse(a, 1000000009LL)) % 1000000009LL, k,
               1000000009LL);
    long long int up = (n + 1) / k;
    if (Y == 1) {
      cout << (X * up) % 1000000009LL << endl;
      return 0;
    }
    long long int temp =
        (((X * (bigMod(Y, up, 1000000009LL) - 1)) % 1000000009LL) *
             (max(1LL, modInverse((Y - 1 + 1000000009LL) % 1000000009LL,
                                  1000000009LL)) %
              1000000009LL) +
         1000000009LL) %
        1000000009LL;
    ans = temp;
    cout << ans << endl;
  }
}

```