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>

#define int long long
#define double long double
#define ff first
#define ss second
#define endl '\n'
#define ii pair<int, int>
#define mp make_pair
#define mt make_tuple
#define DESYNC                      \
  ios_base::sync_with_stdio(false); \
  cin.tie(0);                       \
  cout.tie(0)
#define pb push_back
#define vi vector<int>
#define vii vector<ii>
#define all(x) x.begin(), x.end()
#define EPS 1e-9
#define INF 1e18
#define ROOT 1
#define M 1000000007
#define curtime chrono::steady_clock::now().time_since_epoch().count
#define rep(i, beg, n, s) for (int i = beg; i < n; i += s)
const double PI = acos(-1);

using namespace std;

inline int mod(int n, int m = M) {
  int ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}

int exp(int n, int k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  int ax = exp(n, k / 2);
  ax = mod(ax * ax);
  if (k % 2) ax = mod(ax * n);
  return ax;
}

int gcd(int a, int b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
//#define MULTIPLE_TEST_CASE
void solution() {
  int n, m;
  cin >> n >> m;
  ii v[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> v[i].ff >> v[i].ss;
  }

  set<tuple<int, int, int>> s;
  vector<ii> ans(n + 2, ii(INF, INF));
  for (int i = 1; i <= n; i++) {
    s.insert(mt(0, -i, 1));
  }

  while (s.size() > 0) {
    // climb down
    int dst, y, x;
    tie(dst, y, x) = *s.begin();
    y = -y;
    s.erase(s.begin());
    if (x == n + 1) break;
    int cur = dst;
    if (v[x].ff <= y && y <= v[x].ss) {
      cur += (v[x].ss - y + 1);
      y = v[x].ss + 1;
    }
    if (y > m) {
      continue;
    }
    cur++;
    if (ans[x + 1] > ii(cur, y)) {
      ans[x + 1] = ii(cur, y);
      s.insert(mt(cur, -y, x + 1));
    }
  }
  for (int i = 2; i <= n + 1; i++) {
    cout << (ans[i].ff >= INF ? -1 : ans[i].ff) << endl;
  }
}

int32_t main() {
  DESYNC;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int t = 1;
#ifdef MULTIPLE_TEST_CASE
  cin >> t;
#endif
  while (t--) solution();
}