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++)
int H, W, a, b, f, t, e;
set<pair<int,int>> st;
multiset<int> ms;
int main(){
	scanf("%d%d", &H, &W);
	rep(i,W){
		st.emplace(i,i);
		ms.insert(0);
	}
	rep(i,H){
		scanf("%d%d", &a, &b);
		auto tl = st.lower_bound({a-1,0}), el = st.lower_bound({b,0});
		e = tl != el;
		for(; tl != el; tl = st.erase(tl)){
			tie(t,f) = *tl;
			ms.erase(ms.find(t-f));
		}
		if(e && b != W){
			st.emplace(b,f);
			ms.insert(b-f);
		}
		printf("%d\n", ms.empty() ? -1 : *ms.begin()+i+1);
	}
}