#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000 * 1000 * 1000 + 9;
int mul(long long a, long long b) { return (a * b) % MOD; }
int fpow(int x, int y) {
  int base = x % MOD;
  int ret = 1;
  for (int i = 0; y >> i; i++) {
    if ((y >> i) & 1) ret = mul(ret, base);
    base = mul(base, base);
  }
  return ret;
}
int inv(int x) { return fpow(x, MOD - 2); }
char s[100010];
int main() {
  int n, a, b, k;
  scanf("%d %d %d %d %s", &n, &a, &b, &k, s);
  int c = 0, p = (n + 1) / k;
  for (int i = 0; i < k; i++) {
    int cur = mul(fpow(a, n - i), fpow(b, i));
    if (s[i] == '-') cur = (MOD - cur) % MOD;
    c = (c + cur) % MOD;
  }
  int x = fpow(mul(inv(a), b), k);
  int ans;
  if (x == 1) {
    ans = mul(c, p);
  } else {
    int num = (MOD + fpow(x, p) - 1) % MOD;
    int den = (MOD + x - 1) % MOD;
    ans = mul(c, mul(num, inv(den)));
  }
  printf("%d\n", ans);
  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 MOD = 1000 * 1000 * 1000 + 9;
int mul(long long a, long long b) { return (a * b) % MOD; }
int fpow(int x, int y) {
  int base = x % MOD;
  int ret = 1;
  for (int i = 0; y >> i; i++) {
    if ((y >> i) & 1) ret = mul(ret, base);
    base = mul(base, base);
  }
  return ret;
}
int inv(int x) { return fpow(x, MOD - 2); }
char s[100010];
int main() {
  int n, a, b, k;
  scanf("%d %d %d %d %s", &n, &a, &b, &k, s);
  int c = 0, p = (n + 1) / k;
  for (int i = 0; i < k; i++) {
    int cur = mul(fpow(a, n - i), fpow(b, i));
    if (s[i] == '-') cur = (MOD - cur) % MOD;
    c = (c + cur) % MOD;
  }
  int x = fpow(mul(inv(a), b), k);
  int ans;
  if (x == 1) {
    ans = mul(c, p);
  } else {
    int num = (MOD + fpow(x, p) - 1) % MOD;
    int den = (MOD + x - 1) % MOD;
    ans = mul(c, mul(num, inv(den)));
  }
  printf("%d\n", ans);
  return 0;
}

```