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 main(){
    int h, w;
    cin >> h >> w;

    //map[今] = 開始の最大値
    map<int,int> s;
    //答えを求める用
    multiset<int> vals;

    for(int i=0; i<w; ++i){
        s[i]  = i;
    }

    for(int i=0; i<w; ++i){
        vals.insert(0);
    }

    for(int i=0; i<h; ++i){
        int l, r;
        cin >> l >> r;
        --l;
        //init most right
        int mr = -1;

        //l以上で一番小さい値を返す
        auto it = s.lower_bound(l);
        while(it != s.end() && it->first <= r){
            mr = max(mr, it->second);
            int x = it->first - it->second;
            //削除してインクメントされる
            s.erase(it++);
            vals.erase(vals.find(x));
        }
        //rが画面の端の場合は行けないので　w未満
        if(mr != -1 && r < w){
            vals.insert(r - mr);
            s[r] = mr;
        }

        int ans = -1;
        if(vals.size() > 0){
            //右移動と縦移動の回数
            ans = *vals.begin() + (i+1);
        }
        printf("%d\n", ans);
    }
    return 0;
}