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;
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(0);

	int H, W; cin >> H >> W;
	multiset<int> intervals;
	map<int, int> s;
	for(int i = 0; i < W; i++) intervals.insert(0), s[i] = i;
	int cur = 0;
	for(int i = 0; i < H; i++) {
		int a, b; cin >> a >> b;
		a--;
		auto it = s.lower_bound(a);
		int maxrange = -1;
		while(it != s.end() && it->first <= b) {
			maxrange = max(maxrange, it->second);
			int x = it->first - it->second;
			intervals.erase(intervals.find(x));
			s.erase(it++);
		}
		if(maxrange != -1 && b < W) {
			intervals.insert(b - maxrange);
			s[b] = maxrange;
		}
		int ans = (intervals.empty() ? -1 : *intervals.begin() + i + 1);
		cout << ans << '\n';
	}
}