#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int oo = 0x3f3f3f3f;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline T lowbit(T n) {
  return (n ^ (n - 1)) & n;
}
template <class T>
inline int countbit(T n) {
  return (n == 0) ? 0 : (1 + countbit(n & (n - 1)));
}
template <class T>
inline T gcd(T a, T b) {
  T c;
  while (a != 0) {
    c = a;
    a = b % a;
    b = c;
  }
  return b;
}
template <class T>
inline T mabs(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
typedef vector<int> VI;
typedef vector<VI> VII;
typedef vector<string> VS;
const int N = (int)1e5 + 111;
const long long MOD = (long long)1e9 + 9;
int s[N];
char S[N];
int n, a, b, k;
long long co;
inline long long mpow(long long a, long long b) {
  long long x = 1LL, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    y = (y * y) % MOD;
    b >>= 1LL;
  }
  return x;
}
inline long long getZ(int n, int a, int b, int k, long long q) {
  long long Z = 0LL;
  for (int i = (0); i < int(k); i++) {
    Z += mpow(a, n - i) * mpow(b, i) % MOD * q % MOD * s[i];
  }
  Z = (Z % MOD + MOD) % MOD;
  return Z;
}
inline int solve(int testnum) {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", S);
  for (int i = (0); i < int(k); i++) s[i] = (S[i] == '+' ? 1 : -1);
  if (n < k) k = n + 1;
  long long q = mpow(mpow(a, MOD - 2) * b % MOD, k);
  int numblocks = (n + 1) / k;
  if (q == 1LL)
    q = numblocks;
  else
    q = (mpow(q, n / k + 1) - 1) * mpow(q - 1, MOD - 2) % MOD;
  long long Z = getZ(n, a, b, k, q);
  printf("%lld\n", Z);
  return 0;
}
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  int t = 1;
  if (argc > 1) t = atoi(argv[1]);
  for (int _t = (1); _t < int(t + 1); _t++) {
    int ret = ::solve(_t);
    if (ret == -1) {
      cerr << "Failed test #" << _t << endl;
      break;
    }
    cerr << "Solved test #" << _t << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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:36777216")
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-11;
const int oo = 0x3f3f3f3f;
template <class T>
inline void checkmin(T &a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline T lowbit(T n) {
  return (n ^ (n - 1)) & n;
}
template <class T>
inline int countbit(T n) {
  return (n == 0) ? 0 : (1 + countbit(n & (n - 1)));
}
template <class T>
inline T gcd(T a, T b) {
  T c;
  while (a != 0) {
    c = a;
    a = b % a;
    b = c;
  }
  return b;
}
template <class T>
inline T mabs(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
typedef vector<int> VI;
typedef vector<VI> VII;
typedef vector<string> VS;
const int N = (int)1e5 + 111;
const long long MOD = (long long)1e9 + 9;
int s[N];
char S[N];
int n, a, b, k;
long long co;
inline long long mpow(long long a, long long b) {
  long long x = 1LL, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % MOD;
    }
    y = (y * y) % MOD;
    b >>= 1LL;
  }
  return x;
}
inline long long getZ(int n, int a, int b, int k, long long q) {
  long long Z = 0LL;
  for (int i = (0); i < int(k); i++) {
    Z += mpow(a, n - i) * mpow(b, i) % MOD * q % MOD * s[i];
  }
  Z = (Z % MOD + MOD) % MOD;
  return Z;
}
inline int solve(int testnum) {
  scanf("%d%d%d%d", &n, &a, &b, &k);
  scanf("%s", S);
  for (int i = (0); i < int(k); i++) s[i] = (S[i] == '+' ? 1 : -1);
  if (n < k) k = n + 1;
  long long q = mpow(mpow(a, MOD - 2) * b % MOD, k);
  int numblocks = (n + 1) / k;
  if (q == 1LL)
    q = numblocks;
  else
    q = (mpow(q, n / k + 1) - 1) * mpow(q - 1, MOD - 2) % MOD;
  long long Z = getZ(n, a, b, k, q);
  printf("%lld\n", Z);
  return 0;
}
int main(int argc, char *argv[]) {
  ios::sync_with_stdio(false);
  int t = 1;
  if (argc > 1) t = atoi(argv[1]);
  for (int _t = (1); _t < int(t + 1); _t++) {
    int ret = ::solve(_t);
    if (ret == -1) {
      cerr << "Failed test #" << _t << endl;
      break;
    }
    cerr << "Solved test #" << _t << endl;
  }
  return 0;
}

```