963_A. Alternating Sum_8883
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}.
MOD = 1000000009
def xpow(desk,step):
    ret=1
    desk=desk%MOD
    while step:
        if(step%2):
            ret=(ret*desk)%MOD
        step=step//2
        desk = (desk * desk) % MOD
    return ret
if __name__ == '__main__':
    n,a,b,k=map(int,input().split())
    s=input()
    base=0
    for i in range(0,k):
        base = (base + [-1, 1][s[i] == '+'] *(xpow(a, n - i) * xpow(b, i))) % MOD
    loop=(n+1)//k
    a=xpow(a,k)
    b=xpow(b,k)
    if(a!=b):
        print((((xpow(b,loop)-xpow(a,loop))*xpow((b*xpow(a,loop-1)-xpow(a,loop)),MOD-2))%MOD*base)%MOD)
    else:
        print((base*loop)%MOD)