Problem: There is a grid of squares with H+1 horizontal rows and W vertical columns.

You will start at one of the squares in the top row and repeat moving one square right or down. However, for each integer i from 1 through H, you cannot move down from the A_i-th, (A_i + 1)-th, \ldots, B_i-th squares from the left in the i-th row from the top.

For each integer k from 1 through H, find the minimum number of moves needed to reach one of the squares in the (k+1)-th row from the top. (The starting square can be chosen individually for each case.) If, starting from any square in the top row, none of the squares in the (k+1)-th row can be reached, print `-1` instead.

Constraints

* 1 \leq H,W \leq 2\times 10^5
* 1 \leq A_i \leq B_i \leq W

Input

Input is given from Standard Input in the following format:


H W
A_1 B_1
A_2 B_2
:
A_H B_H


Output

Print H lines. The i-th line should contain the answer for the case k=i.

Example

Input

4 4
2 4
1 1
2 3
2 4


Output

1
3
6
-1

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
#define ll long long
#define lld long double
#define MOD 1000000007
#define inf 1000000000000000000ll
#define pii pair<int,int>
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
ll power(ll x,ll y, ll md=inf){ll res = 1;x%=md;while(y){if(y&1)res = (res*x)%md;x *= x; if(x>=md) x %= md; y >>= 1;}return res;}
 
using namespace std;


#define endl '\n'
#define int ll

signed main() {
	fast;

	int h,w;
	cin>>h>>w;

	vector<pii> v(h);
	for(int i=0;i<h;i++) cin>>v[i].f>>v[i].s;

	map<int,int> mp;
	for(int i=1;i<=w;i++) mp[i] = 0;

	map<int, int> mini = {{0,w}};

	for(int i=0;i<h;i++) {
		int l = v[i].f;
		int r = v[i].s;

		auto it = mp.lower_bound(l);
		int mn = inf;

		while(it!=mp.end() && it->f <= r) {
			mn = min(mn, r+1-(it->f)+(it->s));
			auto it1 = it++;
			mini[it1->s]--;
			if(mini[it1->s]==0) mini.erase(it1->s);
			mp.erase(it1);
		}

		if(r+1<=w && mn<inf) {
			if(mp.count(r+1) && mp[r+1] > mn) {
				mini[mp[r+1]]--;
				if(mini[mp[r+1]]==0) mini.erase(mp[r+1]);
				mini[mn]++;
				mp[r+1] = mn;
			}
			else if(!mp.count(r+1)){
				mp[r+1] = mn;
				mini[mn]++;
			}
		}

		if(mini.size()) cout<<mini.begin()->f+i+1<<endl;
		else cout<<-1<<endl;

	}
}