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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double lf;
typedef pair<int,int> ii;
typedef pair<ii,int> iii;
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define RST(i,n) memset(i,n,sizeof i)
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(),a.end()
#define X first
#define Y second
#define eb emplace_back
#ifdef cold66
#define debug(...) do{\
    fprintf(stderr,"LINE %d: (%s) = ",__LINE__,#__VA_ARGS__);\
    _do(__VA_ARGS__);\
}while(0)
template<typename T>void _do(T &&_x){cerr<<_x<<endl;}
template<typename T,typename ...S> void _do(T &&_x,S &&..._t){cerr<<_x<<", ";_do(_t...);}
template<typename _a,typename _b> ostream& operator << (ostream &_s,const pair<_a,_b> &_p){return _s<<"("<<_p.X<<","<<_p.Y<<")";}
template<typename It> ostream& _OUTC(ostream &_s,It _ita,It _itb)
{
    _s<<"{";
    for(It _it=_ita;_it!=_itb;_it++)
    {
        _s<<(_it==_ita?"":",")<<*_it;
    }
    _s<<"}";
    return _s;
}
template<typename _a> ostream &operator << (ostream &_s,vector<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a> ostream &operator << (ostream &_s,set<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a,typename _b> ostream &operator << (ostream &_s,map<_a,_b> &_c){return _OUTC(_s,ALL(_c));}
template<typename _t> void pary(_t _a,_t _b){_OUTC(cerr,_a,_b);cerr<<endl;}
#define IOS()
#else
#define debug(...)
#define pary(...)
#define endl '\n'
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0);
#endif // cold66
//}
template<class T> inline bool chkmax(T &a, const T &b) { return b > a ? a = b, true : false; }
template<class T> inline bool chkmin(T &a, const T &b) { return b < a ? a = b, true : false; }

const ll MAXn=2e5+5,MAXlg=__lg(MAXn)+2;
const ll MOD=1000000007;
const ll INF=0x3f3f3f3f3f3f3f3f;

ll a[MAXn],b[MAXn];
ll bit[MAXn];
ll lowbit(ll x){return x&(-x);}
void ins(ll x,ll val) {
    for (x++;x<MAXn;x+=lowbit(x)) {
        bit[x] += val;
    }
}
ll qr(ll x){
    ll ret = 0;
    for (x++;x>0;x-=lowbit(x)) {
        ret += bit[x];
    }
    return ret;
}
int main(){
    IOS();
    ll n,m;
    cin >> n >> m;
    for (ll i=1;i<=n;++i) {
        cin >> a[i] >> b[i];
    }
    
    ll cur = 1, ans = 0;
    for (ll i=1;i<=n;++i) {
        if (ans == -1) {
            cout << ans << endl;
            continue;
        }
        if (b[i] < cur) {
            if (ans == -1) cout << ans << endl;
            else {
                ans++;
                cout << ans << endl;
            }
            continue;
        }
        ins(a[i],1), ins(b[i]+1,-1);
        if (qr(cur) > 0) {
            ll tmp = ans + b[i]+1 - cur;
            for (ll j=cur;j<=b[i]+1;j++) {
                if (qr(j) == 0) {
                    tmp = min(tmp,i-1 + b[i]+1 - j);
                }
            }
            ans = tmp;
            cur = b[i]+1;
        }
        ans++;
        if (cur == n+1) ans = -1;
        cout << ans << endl;
    }
}