#include <bits/stdc++.h>
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 9;
int n, a, b, k;
char s[MAXN];
namespace {
inline int add(int x) { return x >= MOD ? x - MOD : x; }
inline int sub(int x) { return x < 0 ? x + MOD : x; }
inline int add(int x, int y) { return add(x + y); }
inline int sub(int x, int y) { return sub(x - y); }
inline int mul(int x, int y) { return (long long)x * y % MOD; }
inline int pwr(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = mul(x, x)) {
    if (y & 1) ans = mul(ans, x);
  }
  return ans;
}
inline int inv(int x) { return pwr(x, MOD - 2); }
}  // namespace
namespace solver1 {
void main() {
  int sum1 = 0;
  for (int i = 0; i < k; i++) {
    int val = mul(inv(pwr(a, i)), pwr(b, i));
    if (s[i] == '+') {
      sum1 = add(sum1, val);
    } else {
      sum1 = sub(sum1, val);
    }
  }
  int sum2 = 0;
  int x = pwr(mul(inv(a), b), k);
  if (x == 1) {
    sum2 = (n + 1) / k;
  } else {
    sum2 = mul(sub(pwr(x, (n + 1) / k), 1), inv(sub(x, 1)));
  }
  printf("%d\n", mul(pwr(a, n), mul(sum1, sum2)));
}
}  // namespace solver1
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", s);
  solver1::main();
}

### Prompt

Generate 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>
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 9;
int n, a, b, k;
char s[MAXN];
namespace {
inline int add(int x) { return x >= MOD ? x - MOD : x; }
inline int sub(int x) { return x < 0 ? x + MOD : x; }
inline int add(int x, int y) { return add(x + y); }
inline int sub(int x, int y) { return sub(x - y); }
inline int mul(int x, int y) { return (long long)x * y % MOD; }
inline int pwr(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = mul(x, x)) {
    if (y & 1) ans = mul(ans, x);
  }
  return ans;
}
inline int inv(int x) { return pwr(x, MOD - 2); }
}  // namespace
namespace solver1 {
void main() {
  int sum1 = 0;
  for (int i = 0; i < k; i++) {
    int val = mul(inv(pwr(a, i)), pwr(b, i));
    if (s[i] == '+') {
      sum1 = add(sum1, val);
    } else {
      sum1 = sub(sum1, val);
    }
  }
  int sum2 = 0;
  int x = pwr(mul(inv(a), b), k);
  if (x == 1) {
    sum2 = (n + 1) / k;
  } else {
    sum2 = mul(sub(pwr(x, (n + 1) / k), 1), inv(sub(x, 1)));
  }
  printf("%d\n", mul(pwr(a, n), mul(sum1, sum2)));
}
}  // namespace solver1
int main() {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", s);
  solver1::main();
}

```