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;

const int MAXN = 200005;
const int MAX = 100000000;

int tree[4 * MAXN], flag[4 * MAXN];

void push(int i, int L, int R){
	if(flag[i] == -1)return;
	
	if(L < R){
		flag[2 * i] = flag[i];
		flag[2 * i + 1] = flag[i];
	}
	
	tree[i] = flag[i] + L;
	flag[i] = -1;
}
void update(int i, int L, int R, int u, int v, int val){
	push(i, L, R);
	
	if(v < L || R < u)return;
	if(u <= L && R <= v){
		flag[i] = val;
		push(i, L, R);
		
		return;
	}
	
	int mid = (L + R) / 2;
	
	update(2 * i, L, mid, u, v, val);
	update(2 * i + 1, mid + 1, R, u, v, val);
	
	tree[i] = min(tree[2 * i], tree[2 * i + 1]);
}
int search(int i, int L, int R, int u){
	push(i, L, R);
	
	if(L == R){
		return tree[i];
	}
	
	int mid = (L + R) / 2;
	if(u <= mid)return search(2 * i, L, mid, u);
	else return search(2 * i + 1, mid + 1, R, u);
}

int h, w, a[MAXN], b[MAXN];

int main(){
	scanf("%d %d", &h, &w);
	
	memset(flag, -1, sizeof flag);
	
	for(int i = 0; i < h; i++){
		scanf("%d %d", &a[i], &b[i]);
		a[i]--, b[i]--;
		
		if(a[i] == 0){
			update(1, 0, w - 1, a[i], b[i], MAX);
		}else{
			int r = search(1, 0, w - 1, a[i] - 1);
			update(1, 0, w - 1, a[i], b[i], r - (a[i] - 1));
		}
		
		if(tree[1] >= MAX)printf("-1\n");
		else printf("%d\n", tree[1] + i + 1);
	}
	
	return 0;
}