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>
#include <string>
#include <vector>
#include <algorithm>
#define rep(i,n) for (int i = 0;i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
int INF = 1e9;
void chmax(int& a, int b) { a = max(a, b);}
void chmin(int& a,int b) {a = min(a,b);}

int main(){
    int h,w;cin >> h >> w;
    map<int,int> mp;
    multiset<int> m;
    rep(i,w) mp[i] = i;
    rep(i,w) m.insert(0);
    rep(i,h){
        int l,r;cin >> l >> r;
        --l;
        map<int,int>::iterator it = mp.lower_bound(l);
        int mr = -1;
        while(it != mp.end() && it->first <= r){
            mr = it->second;
            m.erase(m.find(it->first - it->second));
            mp.erase(it++);
        }
        if(mr != -1 && r < w) {
            mp[r] = mr;
            m.insert(r-mr);
        }
        int  ans = -1;
        if(m.size()>0) {
            ans = *m.begin() + (i+1);
        }
        cout << ans << endl;

    }
}