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 s[800001], la[800001];
void upd(int id, int v){
	la[id] = v, s[id] = v;
}
void prop(int id, int l, int r){
	if (la[id] == 0) return;
	int mid = (l + r) / 2;
	upd(id * 2, la[id]);
	upd(id * 2 + 1, la[id] + mid - l + 1);
	la[id] = 0;
}
void update(int id, int l, int r, int L, int R, int v){
	if (r < L || R < l){
		return;
	}else if (L <= l && r <= R){
		upd(id, v);
		return;
	}
	prop(id, l, r);
	int mid = (l + r) / 2;
	update(id * 2, l, mid, L, R, v);
	update(id * 2 + 1, mid + 1, r, L, R, v + mid - l + 1);
	s[id] = min(s[id * 2], s[id * 2 + 1]);
}
int query(int id, int l, int r, int L, int R){
	if (r < L || R < l){
		return 2147483647;
	}else if (L <= l && r <= R){
		return s[id];
	}
	prop(id, l, r);
	int mid = (l + r) / 2;
	int t1 = query(id * 2, l, mid, L, R);
	int t2 = query(id * 2 + 1, mid + 1, r, L, R);
	return min(t1, t2);
}
int main(){
	int n, m, x, y, l = 0;
	scanf("%d %d", &n, &m);
	for (int i = 0; i < n; i++){
		scanf("%d %d", &x, &y), x--, y--;
		if (x <= l){
			l = max(l, y + 1);
		}else{
			update(1, 0, m - 1, x, y, query(1, 0, m - 1, x - 1, x - 1) + 1 - x);
		}
		if (l == m) printf("-1\n");
		else printf("%d\n", query(1, 0, m - 1, l, m - 1) + i + 1);
	}
}