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>
using namespace std;
#define pii pair<int,int>

const int maxn = 1e6+7;
int t,n,m;
int h,w;
set<pii> nodes; //{ending, starting}
multiset<int> ans;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	cin >> h >> w;
	for (int i=0; i<w; i++){
		nodes.insert({i,i});
		ans.insert(0);
	}
	int x,y;
	for (int i=0; i<h; i++){
		cin >> x >> y;
		--x, --y;
		auto itst = nodes.lower_bound({x,-maxn});
		auto ited = nodes.upper_bound({y+1, -maxn});
		int mx =-1;
		
		while (itst != ited){
			mx = itst->second;
			
			ans.erase(ans.find(itst-> first - mx));
			nodes.erase(itst++); //em.. can't separate mp.erase and it++
		}
		
		if (mx > -1 && y < w-1){
			nodes.insert({y+1, mx});
			ans.insert(y+1-mx);
		}
		
		if (ans.size()) cout << *ans.begin() + i + 1 << endl;
		else cout << -1 << endl;
	}
	return 0;
}