Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

4 4
2 4
1 1
2 3
2 4

Sample Output 1

1
3
6
-1

Let (i,j) denote the square at the i-th row from the top and j-th column from the left.

For k=1, we need one move such as (1,1) → (2,1).

For k=2, we need three moves such as (1,1) → (2,1) → (2,2) → (3,2).

For k=3, we need six moves such as (1,1) → (2,1) → (2,2) → (3,2) → (3,3) → (3,4) → (4,4).

For k=4, it is impossible to reach any square in the fifth row from the top.

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;

#define REP(i, n, m) for(ll i=n; i<(ll)m; ++i)
#define IREP(i, n, m) for(ll i=n-1; i>=m; --i)
#define rep(i, n) REP(i, 0, n)
#define irep(i, n) IREP(i, n, 0)
#define all(v) v.begin(), v.end()
#define vprint(v) for(auto e:v){cout<<e<<" ";};cout<<endl;
#define vvprint(vv) for(auto v:vv)vprint(v);

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(20);

    ll H, W;
    cin >> H >> W;
    vll A(H), B(H);
    rep(i, H) cin >> A[i] >> B[i];

    map<int, int> m;
    rep(i, W) m[i] = 0;
    multiset<int> ms;
    rep(i, W) ms.insert(0);
    vll ans(H, 0);
    rep(i, H){
        int a = A[i]-1;
        int b = B[i]-1;
        auto lb = m.lower_bound(a);
        auto ub = m.upper_bound(b);
        if(lb!=m.end() && lb!=ub){
            --ub;
            auto it = lb;
            int k = ub->first;
            int v = ub->second;
            while(it!=m.end() && it->first!=k){
                int v = it->second;
                ms.erase(ms.find(v));
                it = m.erase(it);
            }
            if(b<W-1 && m.count(b+1)==0){
                m[b+1] = v+b-k+1;
                ms.insert(v+b-k+1);
            }
            m.erase(k);
            ms.erase(ms.find(v));
        }
        if(ms.empty()) ans[i] = -1;
        else ans[i] = *min_element(all(ms)) + i + 1;
    }
    rep(i, H) cout << ans[i] << endl;
}