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 <iostream>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
map<ll,pair<ll,int>> s;
set<pair<ll,int>> t;
ll inf = 10000000000;
ll solve(ll x){
    auto it = s.upper_bound(x);
    it--;
    auto p = *it;
    return p.second.first + p.second.second*(x - p.first);
}

int curve(int x){
    auto it = s.upper_bound(x);
    it--;
    auto p = *it;
    return p.second.second;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int i,h,w; cin >> h >> w;
    s[0] = {inf,0}; t.insert({inf,0});
    s[1] = {0LL,0}; t.insert({0LL,1});
    s[w + 1] = {inf,0}; 
    for(i=1;i<=h;i++){
        int a,b; cin >> a >> b;
        ll hi1 = solve(a - 1),hi2 = solve(b + 1),k = curve(b + 1);
        t.erase({solve(a),a}); s[a] = {hi1 + 1,1},t.insert({hi1 + 1,a});
        t.erase({solve(b + 1),b + 1}); s[b + 1] = {hi2,k},t.insert({hi2,b + 1});
        auto it = s.upper_bound(a);
        while(true){
            auto q = *it;
            if(q.first>b) break;
            t.erase({q.second.first,q.first}); s.erase(it++);
        }
        auto q = *t.begin();
        if(q.first + i<inf) cout << q.first + i << endl;
        else cout << -1 << endl;
    }
}