Problem: 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.

Example

Input

4 4
2 4
1 1
2 3
2 4


Output

1
3
6
-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1001001001;

int main() {
  int h, w;
  cin >> h >> w;

  map<int, int> s;
  multiset<int> vals;

  rep(i, w) s[i] = i;
  rep(i, w) vals.insert(0);

  rep(i, h) {
    int l, r;
    cin >> l >> r;
    --l;
    auto it = s.lower_bound(l);
    int mr = -1;
    while (it != s.end() && it->first <= r) {
      mr = max(mr, it->second);
      int x = it->first - it->second;
      vals.erase(vals.find(x));
      s.erase(it++);
    }
    if (mr != -1 && r < w) {
      vals.insert(r - mr);
      s[r] = mr;
    }
    int ans = -1;
    if (vals.size() > 0) {
      ans = *vals.begin() + (i + 1);
    }
    cout << ans << endl;
  }
  return 0;
}