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;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
 
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define rep1(i, n) for (ll i = 1; i <= (ll)(n); i++)
#define INF 10000000000
#define MOD 1000000007
using ll = long long;
using pint = pair<int,int>;
using pll = pair<ll,ll>;
using Graph = vector<vector<int>>;

int main(){
    int H,W; cin>>H>>W;
    map<int,int> S;//S[now] = start_ind
    multiset<int> vals;//横方向の差分の集合
    //__init__
    rep(i,W)S[i] = i;
    rep(i,W)vals.insert(0);

    rep(i,H){
        int l,r; cin>>l>>r;
        --l;
        auto it = S.lower_bound(l);//obstacle 上 now;
        int mr = -1;
        while(it != S.end() && it->first <= r){
            //削除候補
            mr = max(mr,it->second);//startの候補を削除
            vals.erase(vals.find(it->first-it->second));
            S.erase(it++);
        }
        if(mr != -1 && r < W){
            vals.insert(r-mr);
            S[r] = mr;
        }
        int ans = -1;
        if(vals.size()>0){
            ans = *vals.begin() + (i+1);
        }
        cout<<ans<<endl;
    }
    return 0;
}