#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ii pair<ll,ll>
#define fi first
#define se second
#define endl '\n'
#define puf push_front
#define pof pop_front
#define pub push_back
#define pob pop_back
#define lb lower_bound
#define ub upper_bound
#define rep(x,s,e) for (auto x=s-(s>e);x!=e-(s>e);(s<e?x++:x--))
#define all(x) (x).begin(),(x).end()
#define sz(x) (int) (x).size()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n,k;
int l[100005],r[100005];
set<ii> s;
// we need to check if the intercept of all ranges is null
// find the range exactly after the tile
// then check the range exactly before it?
bool in(ii i){
auto it=s.ub(ii(i.se,-1));
if (it==s.begin()) return false;
return (*prev(it)).se>i.fi;
}
vector<ii> range; //to do greedy
int calc(){
int curr=0;
int ans=0;
for (auto &it:range){
if (curr<=it.fi && !in(it)){
ans++;
curr=it.se;
}
}
return ans;
}
int main(){
cin.tie(0);
cout.tie(0);
cin.sync_with_stdio(false);
cin>>n>>k;
rep(x,0,n) cin>>l[x]>>r[x];
rep(x,0,n) range.pub(ii(l[x],r[x]));
sort(all(range),[](ii i,ii j){
return i.se<j.se;
});
if (calc()<k){
cout<<"-1"<<endl;
return 0;
}
vector<int> ans;
rep(x,0,n){
if (!in(ii(l[x],r[x]))){
s.insert(ii(l[x],r[x]));
//for (auto &it:s) cout<<it.fi<<"_"<<it.se<<" "; cout<<endl;
//cout<<x<<" "<<calc()<<endl;
if (calc()+sz(s)<k){
s.erase(ii(l[x],r[x]));
}
else{
ans.pub(x);
}
}
}
rep(x,0,k) cout<<ans[x]+1<<endl;
}
