#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
int fa[200005], n, m, u[200005], v[200005];
long long w[200005], a[200005], sum[200005];
priority_queue<int, vector<int>, greater<int>> que;
set<pair<int, int>> s[200005];
inline int finds(int p) {
	return (fa[p] == p) ? p : (fa[p] = finds(fa[p]));
}
inline int grand(int l, int r) {
	return rand() % (r - l + 1) + l;
}
set<pair<int, int>>::iterator it;
signed main() {
//	ios::sync_with_stdio(0);
///	cin.tie(0);
//	cout.tie(0);
	srand((unsigned int)time(0));
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		fa[i] = i;
	}
	for (int i = 1; i <= m; i++) {
		cin >> u[i] >> v[i] >> w[i];
		sum[i] = w[i] - a[u[i]] - a[v[i]];
//		cout << sum[i] << "||";
		if (sum[i] <= 0) {
			que.push(i);
			continue;
		}
		int op = sum[i] / 2;
		s[u[i]].insert(make_pair(a[u[i]] + op, i));
		s[v[i]].insert(make_pair(a[v[i]] + sum[i] - op, i));
	}
	while (que.size()) {
		int op = que.top();
		que.pop();
		if (finds(u[op]) == finds(v[op])) continue;
		cout << op << " ";
		int p = finds(u[op]), q = finds(v[op]), uu, vv, ww;
		if (s[p].size() < s[q].size()) swap(p, q);
		for (auto i : s[q]) {
			s[p].insert(i);
		}
		s[q].clear();
		fa[q] = p;
		a[p] += a[q];
		if (s[p].size()) {
			it = s[p].begin();
			while (s[p].size() && (*s[p].begin()).first <= a[p]) {
				op = (*s[p].begin()).second;
				uu = finds(u[op]), vv = finds(v[op]);
				if (uu == vv) {
					ww = w[op] - (*s[p].begin()).first;
					s[p].erase(s[p].begin());
					s[vv].erase(make_pair(ww, op));
					continue;
				}
				if (vv == p) swap(uu, vv);
				ww = w[op] - (*s[p].begin()).first;
//				cout << ww << " " << w[op] << ' ' << (*s[p].begin()).first << " " << a[uu] << " '" << a[vv] << "\n";
				s[p].erase(s[p].begin());
				s[vv].erase(make_pair(ww, op));
				if (a[p] + a[vv] >= w[op]) {
					que.push(op);
				} else {
					sum[op] = w[op] - a[uu] - a[vv];
//					cout << sum[op] << "|" << a[p] + (sum[op] * 3 + 3) / 4 << "##" << w[op] - (a[p] + (sum[op] * 3 + 3) / 4) << "\n";
					s[p].insert(make_pair(a[p] + (sum[op] * 3 + 3) / 4, op));
					s[vv].insert(make_pair(w[op] - (a[p] + (sum[op] * 3 + 3) / 4), op));
				}
			}
		}
	}
}
//复杂度相当于每次启发式合并多做一次？
//应该不会高于n(logn)^2吧,m摆在这肯定到不了n^2肯定到不了n^2肯定到不了n^2肯定到不了n^2肯定到不了n^2肯定到不了n^2肯定到不了n^2
//别被卡常

//所以我还真不知道set erase后迭代器会怎么变化
//cnm，又得想了

//有个乱搞的做法
//可惜有subtask
//没事，信仰随机化,感觉不能太大也不能太小，就取w/4-3w/4

//能过能过能过能过能过能过能过能过能过能过能过能过能过能过能过能过
//能过能过能过能过能过能过能过能过

//好像要是定值
// 直接取3/4,虽然感觉越大越快，但是太大好像也不行