#include <bits/stdc++.h>
//#include <atcoder/modint>

using namespace std;
//using namespace atcoder;
using ll = long long;
//using mint = modint998244353;

struct UnionFind {
    int ngroup, N;
    vector<int> par, siz;

    UnionFind(int _N) : N(_N), par(_N), siz(_N){
        ngroup = _N;
        for(int i = 0; i < N; i++) par[i] = i, siz[i] = 1;
    }

    inline int root(int x) {
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    int unite(int x, int y) {
        int rx = root(x), ry = root(y);
        if (rx == ry) return rx;
        ngroup--;
        if (siz[rx] > siz[ry]) swap(rx, ry);
        par[rx] = ry; siz[ry] += siz[rx];
        return ry;
    }

    inline bool same(int x, int y) {
        int rx = root(x), ry = root(y);
        return rx == ry;
    }

    inline int size(int x) {return siz[root(x)];}
    int group_count() {return ngroup;}

    vector<vector<int>> groups(){
        vector<int> rev(N); int nrt=0;
        for (int i=0; i<N; i++) if (root(i) == i) rev[i] = nrt, nrt++;
        vector<vector<int>> res(nrt);
        for (int i=0; i<N; i++) res[rev[root(i)]].push_back(i);

        return res;
    }
};


int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    /*
       行き来できるペアの数
       sum |S|(|S|-1)/2

       逆から見ていく。
       繋がるごとに差分を計算する。
    */

    ll N, M, K, S=0;
    cin >> N >> M;
    K = N*(N-1)/2;
    vector<ll> u(M), v(M);
    for (int i=0; i<M; i++){
        cin >> u[i] >> v[i];
        u[i]--; v[i]--;
    }
    vector<bool> ng(M);
    ll Q;
    cin >> Q;
    vector<ll> B(Q);
    for (int i=0; i<Q; i++){
        cin >> B[i];
        B[i]--;
        ng[B[i]] = 1;
    }
    UnionFind tree(N);

    vector<ll> ans(Q+1, -1);
    for (int i=0; i<M; i++){
        if (!ng[i]) tree.unite(u[i], v[i]);
    }
    for (int i=0; i<N; i++){
        if (tree.root(i) == i){
            ll sz=tree.size(i);
            S += sz*(sz-1)/2;
        }
    }
    ans[Q] = K-S;
    for (int i=Q-1; i>=0; i--){
        int idx = B[i];
        if (!tree.same(u[idx], v[idx])){
            ll sz1 = tree.size(u[idx]), sz2 = tree.size(v[idx]);
            S -= sz1*(sz1-1)/2 + sz2*(sz2-1)/2;
            sz1 += sz2;
            S += sz1*(sz1-1)/2;
            ans[i] = K-S;
            tree.unite(u[idx], v[idx]);
        }
        else{
            ans[i] = ans[i+1];
        }
    }
    for (int i=1; i<=Q; i++) cout << ans[i] << '\n';

    return 0;
}