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 rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep3(i,m,n) for(int (i)=m;(i)<=(n);(i)++)
#define rep3rev(i,m,n) for(int (i)=m;(i)>=(n);(i)--)
#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define eb emplace_back

using ll = long long;
using vll = vector<ll>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using P = pair<int, int>;

void Main(){
    int h, w; cin >> h >> w;
    map<int, int> mp;
    multiset<int> cost;
    
    rep(i, w) {
        mp[i] = i;
        cost.insert(0);
    }
    
    rep(i, h){
        int a, b;
        cin >> a >> b;
        a--;
        auto it = mp.lower_bound(a);
        int mx = -1;
        while(it != mp.end() && it->fi <= b){
            mx = max(mx, it->se);
            int x = it->fi - it->se;
            cost.erase(cost.find(x));
            mp.erase(it++);
        }
        if(mx != -1 && b < w){
            cost.insert(b - mx);
            mp[b] = mx;
        }
        int ans = -1;
        if(!cost.empty()){
            ans = *cost.begin() + (i+1);
        } 
        cout << ans << endl;
    }
    return;
}

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
    return 0;
}