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 h, w;
map <int, int> available;
multiset <int> minDist;

int main() {
	scanf("%d%d", &h, &w);
	for (int i = 1; i <= w; i++) available[i] = i, minDist.insert(0);
	for (int i = 0; i < h; i++) {
		int a, b; scanf("%d%d", &a, &b);
		auto cur = available.lower_bound(a);
		int l = -1;
		while (cur != available.end() && cur -> first <= b + 1) {
			minDist.erase(minDist.find(cur -> first - cur -> second));
			l = max(l, cur -> second);
			available.erase(cur++);
		}
		if (l != -1 && b < w) {
			available[b + 1] = l;
			minDist.insert(b + 1 - l);
		}
		printf("%d\n", minDist.empty() ? -1 : (*minDist.begin() + i + 1));
	}
	return 0;
}