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)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
#define MOD 998244353
template<class T>inline bool chmax(T& a,T b){if (a<b){a=b;return 1;}return 0;}
template<class T>inline bool chmin(T& a,T b){if (a>b){a=b;return 1;}return 0;}
const long long INF = 1LL << 60;
bool pairCompare(const pair<double,ll>& firstElof, const pair<double,ll>& secondElof){
    return firstElof.first < secondElof.first;
}
bool pairCompareSecond(const pair<double,ll>& firstElof, const pair<double,ll>& secondElof){
    return firstElof.second < secondElof.second;
}
//四方向への移動ベクトル
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

struct edge{//グラフに使うヤツ
    ll from,to,cost;
};

typedef vector<vector<edge> > G;
ll gcd(ll a,ll b){
    if (a%b==0)return(b);
    else return(gcd(b,a%b));
}
int main(){
    ll n,ans=0;
    cin >> n;
    ll a[n];
    ll x[101000];
    fill(x,x+100100,0);
    for (ll i=0;i<n;i++){
        cin >> a[i];
        x[a[i]]=x[a[i]]+1;
        ans+=a[i];
    }
    ll q;
    cin >> q;
    for (ll i=0;i<q;i++){
        ll b,c;
        cin >> b >> c;
        ans+=x[b]*c;
        ans-=x[b]*b;
        x[c]=x[c]+x[b];
        x[b]=0;
        cout << ans << endl;
    }
    return 0;
}