#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
const int MOD = 1e9 + 9;
char s[maxn];
long long n, a, b, k;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y & 1) {
    return x * power(x * x % MOD, y / 2) % MOD;
  } else {
    return power(x * x % MOD, y / 2);
  }
}
int main() {
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  long long niA = power(a, MOD - 2);
  long long niB = power(b, MOD - 2);
  long long tempA = power(a, n);
  long long tempB = 1;
  long long T = 0;
  for (int i = 0; i < k; i++) {
    T = (T + tempA * tempB % MOD * (s[i] == '+' ? 1 : -1) + MOD) % MOD;
    tempA = tempA * niA % MOD;
    tempB = tempB * b % MOD;
  }
  long long d = (n + 1) / k;
  long long fm = (1 - power(1LL * b * niA % MOD, k) + MOD) % MOD;
  long long fz = (1 - power(1LL * b * niA % MOD, (d)*k) + MOD) % MOD;
  if (k == n + 1) {
    cout << T << endl;
  } else if (a == b || fm == 0) {
    cout << T * d % MOD << endl;
  } else
    cout << T * fz % MOD * power(fm, MOD - 2) % MOD << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int maxn = 1e5 + 100;
const int MOD = 1e9 + 9;
char s[maxn];
long long n, a, b, k;
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y & 1) {
    return x * power(x * x % MOD, y / 2) % MOD;
  } else {
    return power(x * x % MOD, y / 2);
  }
}
int main() {
  cin >> n >> a >> b >> k;
  scanf("%s", s);
  long long niA = power(a, MOD - 2);
  long long niB = power(b, MOD - 2);
  long long tempA = power(a, n);
  long long tempB = 1;
  long long T = 0;
  for (int i = 0; i < k; i++) {
    T = (T + tempA * tempB % MOD * (s[i] == '+' ? 1 : -1) + MOD) % MOD;
    tempA = tempA * niA % MOD;
    tempB = tempB * b % MOD;
  }
  long long d = (n + 1) / k;
  long long fm = (1 - power(1LL * b * niA % MOD, k) + MOD) % MOD;
  long long fz = (1 - power(1LL * b * niA % MOD, (d)*k) + MOD) % MOD;
  if (k == n + 1) {
    cout << T << endl;
  } else if (a == b || fm == 0) {
    cout << T * d % MOD << endl;
  } else
    cout << T * fz % MOD * power(fm, MOD - 2) % MOD << endl;
  return 0;
}

```