#include <bits/stdc++.h>
using namespace std;
const int MAXM = 1e6;
const long long MOD = 1e9 + 9;
const long long ep = 1e9 + 5;
const int INF = 1e9;
const double PI = acos(-1);
long long n, a, b, k;
char s[MAXM + 10];
long long acce(long long x, long long y) {
  long long ans = 1;
  long long powe = x;
  long long sum = x;
  long long mul = 1;
  while (y) {
    if (y & 1) {
      ans += mul * sum % MOD;
      ans %= MOD;
      mul *= powe;
      mul %= MOD;
    }
    sum *= (powe + 1);
    sum %= MOD;
    powe *= powe;
    powe %= MOD;
    y /= 2;
  }
  return ans;
}
long long qpow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % MOD;
    x = (x * x) % MOD;
    y = y >> 1;
  }
  return res;
}
long long T(long long m, long long n) {
  if (n <= 1) return 1;
  long long TN2 = T(m, n / 2);
  if (n % 2 == 0) {
    return (TN2 + qpow(m, n / 2) * TN2 % MOD) % MOD;
  } else {
    return (TN2 + qpow(m, n / 2) * TN2 % MOD + qpow(m, n - 1)) % MOD;
  }
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long i = 0;
  long long tp = 0;
  long long tx;
  while (i < k) {
    tx = qpow(a, n - i) * qpow(b, i) % MOD;
    if (s[i] == '+')
      tp = (tp + tx) % MOD;
    else
      tp = (tp - tx + MOD) % MOD;
    i++;
  }
  long long xl = (n + 1) / k;
  long long inv = qpow(a, MOD - 2);
  inv = qpow(inv, k);
  long long ce = qpow(b, k) * inv % MOD;
  long long ans = T(ce, xl);
  ans = (ans * tp) % MOD;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
const int MAXM = 1e6;
const long long MOD = 1e9 + 9;
const long long ep = 1e9 + 5;
const int INF = 1e9;
const double PI = acos(-1);
long long n, a, b, k;
char s[MAXM + 10];
long long acce(long long x, long long y) {
  long long ans = 1;
  long long powe = x;
  long long sum = x;
  long long mul = 1;
  while (y) {
    if (y & 1) {
      ans += mul * sum % MOD;
      ans %= MOD;
      mul *= powe;
      mul %= MOD;
    }
    sum *= (powe + 1);
    sum %= MOD;
    powe *= powe;
    powe %= MOD;
    y /= 2;
  }
  return ans;
}
long long qpow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % MOD;
    x = (x * x) % MOD;
    y = y >> 1;
  }
  return res;
}
long long T(long long m, long long n) {
  if (n <= 1) return 1;
  long long TN2 = T(m, n / 2);
  if (n % 2 == 0) {
    return (TN2 + qpow(m, n / 2) * TN2 % MOD) % MOD;
  } else {
    return (TN2 + qpow(m, n / 2) * TN2 % MOD + qpow(m, n - 1)) % MOD;
  }
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &a, &b, &k);
  scanf("%s", s);
  long long i = 0;
  long long tp = 0;
  long long tx;
  while (i < k) {
    tx = qpow(a, n - i) * qpow(b, i) % MOD;
    if (s[i] == '+')
      tp = (tp + tx) % MOD;
    else
      tp = (tp - tx + MOD) % MOD;
    i++;
  }
  long long xl = (n + 1) / k;
  long long inv = qpow(a, MOD - 2);
  inv = qpow(inv, k);
  long long ce = qpow(b, k) * inv % MOD;
  long long ans = T(ce, xl);
  ans = (ans * tp) % MOD;
  printf("%lld\n", ans);
  return 0;
}

```