#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
constexpr long long infValue =
    std::is_same<T, int>::value ? 2000000007 : 8600000000000000007ll;
template <typename T>
constexpr long long maxValue =
    std::is_same<T, int>::value ? 1000000007 : 1000000000000000001ll;
const long long INF = infValue<long long>;
const long long MOD = 1000000009ll;
const double EPS = 1e-6;
const long long MAX = maxValue<long long>;
long long binPow(long long a, long long x) {
  long long r = 1;
  while (x) {
    if (x & 1) r = r * a % MOD;
    a = a * a % MOD;
    x >>= 1;
  }
  return r;
}
int n, a, b, k;
long long answ;
char s[100101];
int main(void) {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", s);
  for (register long long i = 0; i < k; ++i) {
    answ += (s[i] == '+' ? 1ll : -1ll) * binPow(a, n - i) * binPow(b, i) % MOD;
    answ = (answ + MOD) % MOD;
  }
  n = (n + 1) / k;
  long long t = binPow(b, k) * binPow(binPow(a, k), MOD - 2) % MOD;
  if (t == 1)
    answ = answ * (n + 0) % MOD;
  else
    answ = answ * (1 - binPow(t, n) + MOD) % MOD *
           binPow(1 - t + MOD, MOD - 2) % MOD;
  printf("%lld\n", answ);
  return 0;
}

### Prompt

Create a solution in CPP for 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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
constexpr long long infValue =
    std::is_same<T, int>::value ? 2000000007 : 8600000000000000007ll;
template <typename T>
constexpr long long maxValue =
    std::is_same<T, int>::value ? 1000000007 : 1000000000000000001ll;
const long long INF = infValue<long long>;
const long long MOD = 1000000009ll;
const double EPS = 1e-6;
const long long MAX = maxValue<long long>;
long long binPow(long long a, long long x) {
  long long r = 1;
  while (x) {
    if (x & 1) r = r * a % MOD;
    a = a * a % MOD;
    x >>= 1;
  }
  return r;
}
int n, a, b, k;
long long answ;
char s[100101];
int main(void) {
  scanf("%d %d %d %d", &n, &a, &b, &k);
  scanf("%s", s);
  for (register long long i = 0; i < k; ++i) {
    answ += (s[i] == '+' ? 1ll : -1ll) * binPow(a, n - i) * binPow(b, i) % MOD;
    answ = (answ + MOD) % MOD;
  }
  n = (n + 1) / k;
  long long t = binPow(b, k) * binPow(binPow(a, k), MOD - 2) % MOD;
  if (t == 1)
    answ = answ * (n + 0) % MOD;
  else
    answ = answ * (1 - binPow(t, n) + MOD) % MOD *
           binPow(1 - t + MOD, MOD - 2) % MOD;
  printf("%lld\n", answ);
  return 0;
}

```