注意到已经有两维偏序。
接着数对 $(i,j)$ 有贡献仅当满足以下两组条件之一：

• $i \gt j, T_i \lt T_j, a_i \lt a_j$
• $i \lt j, T_i \lt T_j, a_i \gt a_j$

发现两组条件都有 $T_i \lt T_j$，所以考虑最外层对 $T$ 升序排序。
至于另外两维，就是三维偏序了，分别做一遍 CDQ 即可。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 15;
int n, m;
int p[N], pos[N];
struct node {
    int tm, id, a;
} a[N], b[N];
bool cmp(node a, node b) {
    return a.tm < b.tm;
}

int tr[N];
void add(int x, int d) {
    for ( ; x < N; x += x & -x) tr[x] += d;
}
int query(int x) {
    int res = 0;
    for ( ; x ; x -= x & -x) res += tr[x];
    return res;
}

long long res = 0;
long long sum[N], ans[N];

void init() {
    for (int i = 1; i <= n; i++) {
        sum[i] += (i - 1) - query(p[i]);
        add(p[i], 1);
    }
    for (int i = 1; i <= n; i++) res += sum[i];
    for (int i = 0; i <= n; i++) tr[i] = 0;
    for (int i = n; i >= 1; i--) {
        sum[i] += query(p[i]);
        add(p[i], 1);
    }
    for (int i = 0; i <= n; i++) tr[i] = 0;
}

void cdq1(int l, int r) {
    if (l >= r) return;
    int mid = l + r >> 1;
    cdq1(l, mid), cdq1(mid + 1, r);
    for (int i = l, j = mid + 1, k = l; k <= r; k++) {
        if ((j > r) || (i <= mid && a[i].id > a[j].id)) add(a[i].a, 1), b[k] = a[i++];
        else {
            ans[a[j].tm] += query(a[j].a);
            b[k] = a[j++];
        }
    }
    for (int i = l; i <= mid; i++) add(a[i].a, -1);
    for (int i = l; i <= r; i++) a[i] = b[i];
}
void cdq2(int l, int r) {
    if (l >= r) return;
    int mid = l + r >> 1;
    cdq2(l, mid), cdq2(mid + 1, r);
    for (int i = l, j = mid + 1, k = l; k <= r; k++) {
        if ((j > r) || (i <= mid && a[i].a > a[j].a)) add(a[i].id, 1), b[k] = a[i++];
        else {
            ans[a[j].tm] += query(a[j].id);
            b[k] = a[j++];
        }
    }
    for (int i = l; i <= mid; i++) add(a[i].id, -1);
    for (int i = l; i <= r; i++) a[i] = b[i];
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &p[i]), pos[p[i]] = i;
    for (int i = 1, x; i <= m; i++) scanf("%d", &x), a[i] = (node){i, pos[x], x};
    init();
    sort(a + 1, a + 1 + m, cmp);
    cdq1(1, m);
    sort(a + 1, a + 1 + m, cmp);
    cdq2(1, m);
    sort(a + 1, a + 1 + m, cmp);
    for (int i = 1; i <= m; i++) {
        printf("%lld\n", res);
        res += ans[i] - sum[pos[a[i].a]];
    }
    return 0;
}