964_C. Alternating Sum
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}.
```python
n, a, b, k = map(int, input().split() )
s = input()

mod = (int)( 1e9 + 9 )

ans = 0
t = (int)( (n + 1) / k )

def modInverse(a):
    return pow(a, mod - 2, mod) 

i = 0
for ind in range( len(s) ):
    
    first = ( pow(a, n - i, mod ) * pow( b, i, mod ) ) % mod
    num = ( pow( b, k, mod ) * modInverse( pow(a, k, mod) ) ) % mod
    
    if ( b >= a ) :
        den = ( num - 1 + mod ) % mod
        num = ( pow( num, t, mod ) - 1 + mod ) % mod
    else:
       den = ( 1 - num + mod ) % mod
       num = ( 1 - pow( num, t, mod ) + mod ) % mod
    
    if num == 0:
        temp = ( t * first ) % mod
    else:
        temp = ( num * modInverse(den) ) % mod
        temp = ( temp * first ) % mod
    
    if s[ind] == '+':
        ans = ( ans + temp )%mod
    elif s[ind] == '-':
        ans = ( ans - temp + mod ) % mod
    
    i += 1
    if ( i > n ):
        i = 0

print (ans)
```