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<iostream>
#include<map>
#include<set>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 200005;
int n, m;
int main()
{
	scanf("%d%d", &n, &m);
	map<int, int> mp;
	multiset<int> v;
	for(int i = 1; i <= m; ++i)
	{
		mp[i] = i;
		v.insert(0);
	}

	for(int i = 1, l, r; i <= n; ++i)
	{
		scanf("%d%d", &l, &r);
		auto it = mp.lower_bound(l);
		int newst = 0;
		while(it != mp.end() && it -> first <= r + 1)
		{
			newst = max(newst, it -> second);
			v.erase(v.find(it->first - it->second));
			it = mp.erase(it);
		}
		if(newst && r < m)
		{
			mp[r + 1] = newst;
			v.insert(r + 1 - newst);
		}
		if(v.size())
			printf("%d\n", *v.begin() + i);
		else puts("-1");
	}
	return 0;
}