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; int main(){ int H, W; cin >> H >> W; map<int, int64_t> mp; multiset<int64_t> values; for(int i=1; i<=W; i++){ mp[i] = 0; values.insert(0); }for(int i=1; i<=H; i++){ int l, r; cin >> l >> r; if(r != W){ auto it = mp.upper_bound(r+1); if(it != mp.begin()){ it--; if(it->first != r+1){ int64_t v = it->second + (r+1-it->first); mp[r+1] = v; values.insert(v); } } }auto it = mp.lower_bound(l); while(it != mp.end() && it->first <= r){ values.erase(values.find(it->second)); it = mp.erase(it); }if(values.size()){ cout << i + *values.begin() << endl; }else{cout << -1 << endl; } }return 0; }