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 all(x) (x).begin(),(x).end()
#define rep(i, n) for (int i = 0; i < (n); i++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define endl "\n"
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) {os << "["; for (const auto &v : vec) {os << v << ","; } os << "]"; return os;}
template <typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &p) {os << "(" << p.first << ", " << p.second << ")"; return os;}


void solve() {
    int H, W;
    cin >> H >> W;
    const int inf = 1e9;

    map<int, int> mp; // mp[今] = 最初の位置
    multiset<int> diff; // mpで管理してる区間における 今 - 最初 の集合

    for (int i = 0; i < W; i++) {
        mp[i] = i;
        diff.insert(0);     
    }
    
    for (int i = 0; i < H; i++) {
        // cerr << "mp: ";
        // for(auto e : mp) {
        //     cerr << e << " ";
        // }
        // cerr << endl;

        // cerr << "diff: ";
        // for(auto e : diff) {
        //     cerr << e << " ";
        // }
        // cerr << endl;

        int a, b;
        cin >> a >> b;
        a--; // 半開区間 [a, b)
        auto it = mp.lower_bound(a);
        while(it->first < b && it != mp.end()) {
            if (mp.find(b) == mp.end() || it->second > mp[b]) {
                if (b == W) {
                    if (diff.find(inf) == diff.end()) {
                        diff.insert(inf);
                    }
                } else {
                    if (diff.find(b - mp[b]) != diff.end()) {
                        diff.erase(diff.lower_bound(b - mp[b]));
                    }
                    diff.insert(b - it->second);
                }
                mp[b] = it->second;
            }
            diff.erase(diff.lower_bound(it->first - it->second));
            it = mp.erase(it);
        }

        if (*diff.begin() >= inf) {
            cout << -1 << endl;
        } else {
            cout << *diff.begin() + i + 1 << endl;
        }
    }
    // cerr << "mp: ";
    // for(auto e : mp) {
    //     cerr << e << " ";
    // }
    // cerr << endl;
    // cerr << "diff: ";
    // for(auto e : diff) {
    //     cerr << e << " ";
    // }
    // cerr << endl;

}

int main() {
    #ifdef LOCAL_ENV
    cin.exceptions(ios::failbit);
    #endif
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(16);
    
    solve();
}