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;
typedef long long ll;
const int N = 2e5 + 10;
int h, w, ans, a[N], b[N], cnt[N];
map<int, int> mp;
int main()
{
    std::ios::sync_with_stdio(false);
	cin >> h >> w;
	for(int i = 1; i <= h; i++) cin >> a[i] >> b[i];
	for(int i = 0; i < w; i++) mp[i] = i;
	cnt[0] = w;
	for(int i = 1; i <= h; i++)
	{
		a[i]--;
		b[i]--;
		auto it = mp.lower_bound(a[i]);
		int pos = -1;
		while(it != mp.end() and it -> first <= b[i] + 1)
		{
			pos = max(pos, it -> second);
			cnt[it -> first - it -> second]--;
			it = mp.erase(it);
		}
		if(pos >= 0 and b[i] + 1 < w)
		{
			mp[b[i] + 1] = pos;
			cnt[b[i] + 1 - pos]++;
		}
		while(!cnt[ans] and ans < w) ans++;
		if(ans == w) cout << -1 << endl;
		else cout << i + ans << endl;
	}
	return 0;
}