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;
using ll = long long;

int main() {
	cin.tie(nullptr);
	ios::sync_with_stdio(false);

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

	map<int,int> etos;
	multiset<int> distset;
	for (int i = 0; i < w; i++) {
		etos[i] = i;
		distset.insert(0);
	}
	for (int i = 0; i < h; i++) {
		int a,b;
		cin>>a>>b;
		a--;
		b--;
		int max_s = -1;
		{
			auto bgn = etos.lower_bound(a);
			auto end = etos.upper_bound(b+1);
			for (auto it = bgn; it != end; ++it) {
				max_s = max(max_s, it->second);
				auto dist = it->first - it->second;
				assert(distset.count(dist)>0);
				distset.erase(distset.find(dist));
			}
			etos.erase(bgn, end);
		}
		if (max_s != -1 and b+1 < w) {
			auto min_dist = b+1 - max_s;
			distset.insert(min_dist);
			assert(etos.count(b+1) == 0);
			etos[b+1] = max_s;
		}
		if (!distset.empty()) {
			cout << i+1 + *distset.begin() << endl;
		} else {
			cout << -1 << endl;
		}
	}

	return 0;
}