#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define MAXN (1000005)
ll N,Q;
struct node {
ll s,e,m,lazy;
pair<ll,ll> v;
node *l, *r;
node(ll _s,ll _e){
s = _s;
e = _e;
m = (s + e) / 2;
lazy = -1;
v = {-1,s};
if(s != e){
l = new node(s,m);
r = new node(m + 1,e);
}
}
void propo() {
if(lazy == -1) return;
v.first = max(v.first,lazy);
if(s != e){
l->lazy = max(l->lazy,lazy);
r->lazy = max(r->lazy,lazy);
}
lazy = -1;
}
void update(ll x,ll y,ll nval){
propo();
if(s == x && e == y){
lazy = max(lazy,nval);
return;
}else{
if(x > m) r -> update(x,y,nval);
else if(y <= m) l -> update(x,y,nval);
else l -> update(x,m,nval), r -> update(m + 1,y,nval);
l->propo(), r->propo();
v = min(l->v,r->v);
}
}
pair<ll,ll> rmq(ll x,ll y){
propo();
if(s == x && e == y){
return v;
}else{
if(x > m) return r -> rmq(x,y);
else if(y <= m) return l -> rmq(x,y);
else return min(l -> rmq(x,m),r -> rmq(m + 1,y));
}
}
} *root;
int main() {
ios_base::sync_with_stdio(false);cin.tie(0);
cin>>N>>Q;
root = new node(0,N - 1);
vector<pair<ll,ll> > ranges[N]; //ranges provided with RMQ = i
for(ll q = 0;q < Q;q++){
ll L,R,X;
cin>>L>>R>>X; //0-indexed
root -> update(L,R,X);
ranges[X].push_back({L,R});
}
ll ans[N];
memset(ans,-1,sizeof(ans));
bool ok = 1;
set<ll> values;
for(ll i = 0;i < N;i++){
values.insert(i);
}
for(ll rmq = N - 1;rmq >= 0;rmq--){
if(ranges[rmq].empty()) continue;
ll Lb = -1e18, Ub = 1e18;
for(auto u : ranges[rmq]){
Lb = max(Lb,u.first);
Ub = min(Ub,u.second);
}
if(Lb > Ub){ //no intersection at all between ranges with RMQ = value rmq
ok = 0;
break;
}
pair<ll,ll> a = root -> rmq(Lb,Ub);
if(a.first > rmq){
ok = 0;
}else{
ans[a.second] = rmq; //I can prove that setting any element whose v.first == rmq to value rmq between [Lb,Ub] intersection are always optimal
values.erase(values.find(rmq));
root -> update(a.second,a.second,1e9); //position a.second is already filled up
}
}
vector<pair<ll,ll> > left;
for(ll i = 0;i < N && ok;i++){
if(ans[i] == -1){
pair<ll,ll> a = root -> rmq(i,i);
left.push_back({a.first,i});
}
}
sort(left.begin(),left.end(),greater<pair<ll,ll> >());
for(auto u : left){
if(values.empty() || *prev(values.end()) < u.first){
ok = 0;
break;
}else{
ans[u.second] = *prev(values.end());
values.erase(prev(values.end()));
}
}
if(!ok){
memset(ans,-1,sizeof(ans));
for(ll i = 0;i < N;i++){
cout<<ans[i]<<" ";
}
cout<<'\n';
}else{
for(ll i = 0;i < N;i++){
cout<<ans[i]<<" ";
}
cout<<'\n';
}
}
