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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;


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;
                map<int, int>::iterator 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;
}