#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
char s[N];
int n, k, len;
long long ans, a, b, pt, na, now;
vector<vector<long long> > ori(1, vector<long long>(2)),
    orip(1, vector<long long>(2)), mart(2, vector<long long>(2)),
    martp(2, vector<long long>(2));
void init(long long t, long long a) {
  orip[0][1] = t;
  martp[0][0] = martp[1][0] = 1;
  martp[1][1] = a;
  return;
}
vector<vector<long long> > mul(const vector<vector<long long> > &a,
                               const vector<vector<long long> > &b) {
  vector<vector<long long> > c(a.size(), vector<long long>(b[0].size()));
  for (int i = 0; i < a.size(); i++)
    for (int j = 0; j < b[0].size(); j++)
      for (int k = 0; k < b.size(); k++) {
        c[i][j] = (c[i][j] % 1000000009 + a[i][k] * b[k][j] % 1000000009) %
                  1000000009;
      }
  return c;
}
vector<vector<long long> > quick_pow(vector<vector<long long> > a,
                                     long long n) {
  int sized = a.size();
  vector<vector<long long> > res(sized, vector<long long>(sized));
  for (int i = 0; i < sized; i++) res[i][i] = 1;
  while (n) {
    if (n & 1) {
      res = mul(res, a);
    }
    n >>= 1;
    a = mul(a, a);
  }
  return res;
}
long long quickpow(long long a, long long x) {
  long long ans = 1;
  a %= 1000000009;
  while (x) {
    if (x & 1) ans = (ans * a) % 1000000009;
    x >>= 1;
    a = (a * a) % 1000000009;
  }
  return ans;
}
int main() {
  scanf("%d%I64d%I64d%d", &n, &a, &b, &k);
  ans = 0;
  pt = 1;
  scanf("%s", s);
  pt = quickpow(a, (long long)n);
  ans = ((s[0] == '-' ? -pt : pt) % 1000000009 + 1000000009) % 1000000009;
  len = strlen(s);
  na = quickpow(a, 1000000009 - 2) % 1000000009;
  for (int i = 1; i < len; i++) {
    pt = pt * b % 1000000009 * na % 1000000009;
    ans = (ans +
           ((s[i] == '-' ? -pt : pt) % 1000000009 + 1000000009) % 1000000009) %
          1000000009;
  }
  now = quickpow(na, (long long)k) % 1000000009 * quickpow(b, (long long)k) %
        1000000009;
  len = (n + 1) / k;
  init(ans, now);
  ori = orip;
  mart = martp;
  mart = quick_pow(mart, len);
  ori = mul(ori, mart);
  printf("%I64d\n", ori[0][0]);
  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 N = 1e5 + 10;
char s[N];
int n, k, len;
long long ans, a, b, pt, na, now;
vector<vector<long long> > ori(1, vector<long long>(2)),
    orip(1, vector<long long>(2)), mart(2, vector<long long>(2)),
    martp(2, vector<long long>(2));
void init(long long t, long long a) {
  orip[0][1] = t;
  martp[0][0] = martp[1][0] = 1;
  martp[1][1] = a;
  return;
}
vector<vector<long long> > mul(const vector<vector<long long> > &a,
                               const vector<vector<long long> > &b) {
  vector<vector<long long> > c(a.size(), vector<long long>(b[0].size()));
  for (int i = 0; i < a.size(); i++)
    for (int j = 0; j < b[0].size(); j++)
      for (int k = 0; k < b.size(); k++) {
        c[i][j] = (c[i][j] % 1000000009 + a[i][k] * b[k][j] % 1000000009) %
                  1000000009;
      }
  return c;
}
vector<vector<long long> > quick_pow(vector<vector<long long> > a,
                                     long long n) {
  int sized = a.size();
  vector<vector<long long> > res(sized, vector<long long>(sized));
  for (int i = 0; i < sized; i++) res[i][i] = 1;
  while (n) {
    if (n & 1) {
      res = mul(res, a);
    }
    n >>= 1;
    a = mul(a, a);
  }
  return res;
}
long long quickpow(long long a, long long x) {
  long long ans = 1;
  a %= 1000000009;
  while (x) {
    if (x & 1) ans = (ans * a) % 1000000009;
    x >>= 1;
    a = (a * a) % 1000000009;
  }
  return ans;
}
int main() {
  scanf("%d%I64d%I64d%d", &n, &a, &b, &k);
  ans = 0;
  pt = 1;
  scanf("%s", s);
  pt = quickpow(a, (long long)n);
  ans = ((s[0] == '-' ? -pt : pt) % 1000000009 + 1000000009) % 1000000009;
  len = strlen(s);
  na = quickpow(a, 1000000009 - 2) % 1000000009;
  for (int i = 1; i < len; i++) {
    pt = pt * b % 1000000009 * na % 1000000009;
    ans = (ans +
           ((s[i] == '-' ? -pt : pt) % 1000000009 + 1000000009) % 1000000009) %
          1000000009;
  }
  now = quickpow(na, (long long)k) % 1000000009 * quickpow(b, (long long)k) %
        1000000009;
  len = (n + 1) / k;
  init(ans, now);
  ori = orip;
  mart = martp;
  mart = quick_pow(mart, len);
  ori = mul(ori, mart);
  printf("%I64d\n", ori[0][0]);
  return 0;
}

```