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;
#pragma GCC optimize("O0")
typedef long long int ll;
typedef long double ld;
#define rep(i,n) for (ll i = 0; i < (n); ++i)
#define Rep(i,a,n) for (ll i = (a); i < (n); ++i)
const ll mod = 1e9+7;
const ll INF = 1e18;
#define All(a) (a).begin(),(a).end()
#define Pi acos(-1)
using Graph = vector<vector<ll>>;

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << setprecision(15) << fixed;
  chrono::system_clock::time_point start,end;
  start = chrono::system_clock::now();

  ll h, w;
  cin >> h >> w;
  ll now = 1;
  rep(i,h) {
    ll a, b;
    cin >> a >> b;
    if (now < a) {
      cout << now+i  << '\n';
    }
    else {
      if (now < b+1) now = b+1;
      if (now > w) {
        cout << -1 << '\n';
      }
      else {
        cout << now + i << '\n';
      }
    }
  }

  end = chrono::system_clock::now();
  auto elapsed = chrono::duration_cast< chrono::milliseconds >(end - start).count();
  cerr << elapsed << "ms" << '\n';
}