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 <iostream>
#include <vector>
#include <map>
#include <set>
#define rep(i, n) for(int i=0; i<n; i++)
using namespace std;

const int MAX_H = 200000;
int H, W;
vector<int> A(MAX_H), B(MAX_H);

int main() {
  cin >> H >> W;
  rep(i, H) {
   cin >> A[i] >> B[i];
  }
  
  // map[現在位置] = 開始位置
  map<int, int> mp;
  rep(i, W) mp[i] = i;
  // (現在位置-開始位置）を保持
  multiset<int> st;
  rep(i, W) st.insert(0);
  
  rep(i, H) {
    int l = A[i];
    l--;
    int r = B[i];
    auto it = mp.lower_bound(l);
    int mr = -1;
    while (it != mp.end() && it->first <= r) {
      mr = max(mr, it->second);
      // 下にいけないルートは消す
      int x = it->first - it->second;
      st.erase(st.find(x));
      mp.erase(it++);
    }
    if (mr != -1 && r < W) {
      st.insert(r - mr);
      mp[r] = mr;
    }
    
    // 最短
    int ans = -1;
    if (st.size() > 0) {
      ans = *st.begin() + (i + 1);
    }
    cout << ans << endl;
  }
}