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 int long long

#define ll long long
#define rep(i,n) for(ll i = 0; i < (n); i++)
#define P pair<ll,ll>
#define ld long double
ll INF = (1LL << 60);
int mod = 1000000007;

ll H, W;


signed main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cin >> H >> W;
	// first -> 終点, second -> 始点
	map<int, int>now_place;
	multiset<int>diff;
	rep(i, W) {
		now_place[i] = i;
		diff.insert(0);
	}
	rep(i, H) {
		int A, B; cin >> A >> B; A--;
		auto itr = now_place.lower_bound(A);
		int ma = -1;
		while (itr != now_place.end() && itr->first <= B) {
			ma = max(ma, itr->second);
			int x = itr->first - itr->second;
			diff.erase(diff.find(x));
			now_place.erase(itr++); 
		}
		if (ma != -1 && B < W) {
			diff.insert(B - ma);
			now_place[B] = ma;
		}
		int ans = -1;
		if (diff.size() > 0) {
			ans = i + 1 + *diff.begin();
		}
		cout << ans << endl;
	}
	return 0;
}