You have a sequence A composed of N positive integers: A_{1}, A_{2}, \cdots, A_{N}.

You will now successively do the following Q operations:

* In the i-th operation, you replace every element whose value is B_{i} with C_{i}.



For each i (1 \leq i \leq Q), find S_{i}: the sum of all elements in A just after the i-th operation.

Constraints

* All values in input are integers.
* 1 \leq N, Q, A_{i}, B_{i}, C_{i} \leq 10^{5}
* B_{i} \neq C_{i}

Input

Input is given from Standard Input in the following format:


N
A_{1} A_{2} \cdots A_{N}
Q
B_{1} C_{1}
B_{2} C_{2}
\vdots
B_{Q} C_{Q}


Output

Print Q integers S_{i} to Standard Output in the following format:


S_{1}
S_{2}
\vdots
S_{Q}


Note that S_{i} may not fit into a 32-bit integer.

Examples

Input

4
1 2 3 4
3
1 2
3 4
2 4


Output

11
12
16


Input

4
1 1 1 1
3
1 2
2 1
3 5


Output

8
4
4


Input

2
1 2
3
1 100
2 100
100 1000


Output

102
200
2000
mod = int(1e9+7)
def add(a, b):
    c = a + b
    if c >= mod:
        c -= mod
    return c

def main():
    #A, B = [int(x) for x in raw_input().split()]
    n = int(raw_input())
    
    arr = [int(x) for x in raw_input().split()]
    
    cnt = {}
    
    cur_sum = 0
    
    for num in arr:
        cnt[num] = cnt.get(num, 0) + 1
        cur_sum += num 
    
    q = int(raw_input())
    
    for _ in range(q):
        frm, to = [int(x) for x in raw_input().split()]
        
        frm_cnt, to_cnt = cnt.get(frm, 0), cnt.get(to, 0)
        
        diff = to - frm
        
        total_diff = frm_cnt * diff
        
        cur_sum += total_diff
        
        cnt[to] = to_cnt + frm_cnt
        
        cnt[frm] = 0
        
        print(cur_sum)
        
        
           
        
    
    #queries = [ for x in raw_input().split()]
    

main()