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() {
	int H, W;
	cin >> H >> W;
	map<int, int>s;
	multiset<int>values;

	for (int i = 0; i < W; ++i) {
		s[i] = i;
	}
	for (int i = 0; i < W; ++i) {
		values.insert(0);
	}
	for (int i = 0; i < H; ++i) {
		int l, r;
		cin >> l >> r;
		l--;
		auto it = s.lower_bound(l);
		int mr = -1;
		while (it != s.end() && it->first <= r) {
			mr = max(mr, it->second);
			int x = it->first - it->second;
			values.erase(values.find(x));
			s.erase(it++);
		}
		if ((-1 != mr) && (r < W)) {
			values.insert(r - mr);
			s[r] = mr;
		}
		if (0 == values.size()) {
			cout << -1 <<endl;
		}
		else {
			cout << *values.begin() + (i + 1) << endl;
		}
	}
	return 0;
}