Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

11
12
16

Initially, the sequence A is 1,2,3,4.

After each operation, it becomes the following:

2, 2, 3, 4

2, 2, 4, 4

4, 4, 4, 4

Sample Input 2

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

Sample Output 2

8
4
4

Note that the sequence A may not contain an element whose value is B_{i}.

Sample Input 3

2
1 2
3
1 100
2 100
100 1000

Sample Output 3

102
200
2000

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)


/**
 *    Coded by : lucky_21
 *               --------Lokesh Singh
**/

#include<bits/stdc++.h>
using namespace std;

#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T> using oset=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;

#define     F           first
#define     S           second
#define     pb          push_back
#define     lb          lower_bound
#define     ub          upper_bound
#define     pii         pair<int,int>
#define     all(x)      x.begin(),x.end()
#define     fix         fixed<<setprecision(10)
#define     rep(i,a,b)  for(int i=int(a);i<=int(b);i++)
#define     repb(i,b,a) for(int i=int(b);i>=int(a);i--)
#define     FastIO      ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)

typedef double db;
typedef long long ll;

const int N=1e5+5;
const int mod=1e9+7;

ll n,q,f[N];
signed main(){
    FastIO;
    cin>>n;
    ll sum=0;
    rep(i,1,n){
        int x;
        cin>>x;
        sum+=x;
        f[x]++;
    }
    cin>>q;
    while(q--){
        int a,b;
        cin>>a>>b;
        sum-=a*f[a];
        sum-=b*f[b];
        f[b]+=f[a];
        f[a]=0;
        sum+=b*f[b];
        cout<<sum<<'\n';
    }
//    ll n;
//    cin>>n;
//    string s;
//    while(n){
//        int m=(n%26?n%26:26);
//        m--;
//        s+=char(m+'a');
//        n/=26;
//    }
//    reverse(all(s));
//    cout<<s;
    return 0;
}