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: Wrong Answer)

#pragma GCC optimize("O3")
#pragma GCC target("avx")
#include <stdio.h>
#include <stdlib.h>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) > 0 ? (a) : (-1 * (a)))
#define rep(i, n) for ((i) = 0; (i) < (n); (i)++)
#define rep2(i, s, e) for ((i) = (s); (i) < (e); (i)++)

void marge(int *, int);

int a[200010], l1[200010], l2[200010];
long long main(void)
{

    long long i, j, k, n, q, s = 0, b, c, index = 0;

    scanf("%lld", &n);

    rep(i, n)
    {
        scanf("%lld", a + i);
        s += a[i];
    }
    scanf("%lld", &q);
    marge(a, n);
    l1[0] = a[0];
    l2[0] = 1;
    rep2(i, 1, n)
    {
        if (a[i] == l1[index])
        {
            l2[index] += 1;
        }
        else
        {
            index++;
            l1[index] = a[i];
            l2[index] = 1;
        }
    }

    rep(k, q)
    {
        scanf("%lld %lld", &b, &c);

        i = 0;
        while (i < index + 2)
        {
            if (b == l1[i])
                break;
            i++;
        }
        if (i > index)
        {
            printf("%d\n", s);
            continue;
        }

        j = 0;
        while (j < index + 2)
        {
            if (c == l1[j])
                break;
            j++;
        }
        if (j > index)
        {
            index = j;
            l1[index] = c;
        }
        s += (c - b) * l2[i];
        l2[j] += l2[i];
        l2[i] = 0;
        printf("%d\n", s);
    }

    return 0;
}

void marge(int *a, int n)
{

    if (n == 1)
        return;

    int *tmp;
    tmp = (int *)malloc((n) * sizeof(int));
    marge(a, n / 2);
    marge(a + n / 2, n - n / 2);

    for (int i = 0; i < n / 2; i++)
        tmp[i] = a[i];
    for (int i = n / 2; i < n; i++)
        tmp[i] = a[n / 2 + n - i - 1];

    int lp = 0, rp = n - 1, p = 0;
    int l3, r3;
    while (p < n)
    {
        l3 = tmp[lp];
        r3 = tmp[rp];
        if (l3 < r3)
        {
            a[p] = l3;
            lp++;
            p++;
        }
        else
        {
            a[p] = r3;
            rp--;
            p++;
        }
    }

    return;
}