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;
#define __ <<" "<<
#define ___ <<" "
#define bash push_back
#define ALL(x) x.begin(),x.end()
// #define int long long
struct IoSetup {
    IoSetup() {
        cin.tie(0);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(10);
        cerr << fixed << setprecision(10);
    }
}IoSetup;

using Int = __int128_t;
using ll = long long;
using pii = pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr int SMOD = 1000000007;
constexpr int NMOD = 998244353;
constexpr int dx[]={1,0,-1,0,1,1,-1,-1};
constexpr int dy[]={0,-1,0,1,-1,1,-1,1};

inline bool inside(int x,int y,int w,int h){return (x>=0 && y>=0 && x<w && y<h);}
template<class T>bool chmax(T &a, const T&b){if(a<b)return(a=b,1);return 0;}
template<class T>bool chmin(T &a, const T&b){if(b<a)return(a=b,1);return 0;}

signed main() {

    int h, w;
    cin >> h >> w;

    vector<int> a(h), b(h), ml(h + 1), mr(h + 1);

    ml[0] = 1;
    mr[0] = w;
    bool f = true;
    for(int i = 0; i < h; i++) {
        if(!f) {
            cout << -1 << endl;
            continue;
        }
        cin >> a[i] >> b[i];
        ml[i+1] = ml[i];
        if(a[i] <= ml[i]) {
            ml[i+1] = b[i]+1;
        }
        mr[i+1] = mr[i];
        if(b[i] <= mr[i]) {
            mr[i+1] = a[i] - 1;
        }
        
        if(b[i] == w and a[i] <= ml[i+1]) {
            f = false;
            cout << -1 << endl;
        }
        else {
            cout << i __ ml[i+1] __ mr[i] << endl;
            cout << max(0, ml[i+1] - max(1, mr[i])) + i + 1 << endl;
        }
    }

    return 0;
}