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;
#define rep(i,n) for(int i=0; i<(n); i++)
using P = pair<int,int>;
constexpr int INF = 1e9+7;
priority_queue<P, vector<P>, greater<P>> qu;
int ds[200005], H, W, a, b;
set<P> lv;
int main(){
	scanf("%d%d", &H, &W);
	rep(i,W){
		lv.emplace(i,i);
		qu.emplace(0,i);
	}
	rep(i,H){
		scanf("%d%d", &a, &b);
		auto tl = lv.lower_bound({a-1,0}), el = lv.lower_bound({b,0});
		if(tl != el){
			for(el--; tl != el; tl = lv.erase(tl)) ds[tl->second] = INF;
			auto[to, from] = *el;
			el = lv.erase(el);
			if(b != W){
				ds[from] = b - from;
				lv.emplace_hint(el, b, from);
				qu.emplace(b-from, from);
			}else ds[from] = INF;
		}
		while(!qu.empty()){
			auto [d, from] = qu.top();
			if(ds[from] != d) qu.pop();
			else break;
		}
		printf("%d\n", qu.empty() ? -1 : qu.top().first+i+1);
	}
}