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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
	int h,w;
	cin >> h >> w;

	map<int,int> s;
	multiset<int> vals;
	rep(i,w) s[i] = i;
	rep(i,w) vals.insert(0);

	rep(i,h) {
		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;
			vals.erase(vals.find(x));
			s.erase(it++);
		}
		if (mr != -1 && r < w) {
			vals.insert(r - mr);
			s[r] = mr;
		}
		int ans = -1;
		if (vals.size() > 0) {
			ans = *vals.begin() + (i+1);
		}
		printf("%d\n",ans);
	}
	return 0;
}