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, l, r;
    cin >> h >> w;
    multiset<int> ms; // end - start
    map<int, int> s; // ending, starting
    for(int i = 1; i <= w; i++) { s.insert({i, i}); ms.insert(0); }
    for(int i = 1; i <= h; i++) {
        cin >> l >> r;
        vector<int> to_remove;
		auto it = s.lower_bound(l);
        for(; it != s.end() && (it->first) <= r; it++) {
            to_remove.push_back(it->first);
        }
        int last = -1;
        for(auto x : to_remove) {
            ms.erase(ms.find(x - s[x]));
            last = s[x];
            s.erase(x);
        }
        if(r < w && s.find(r+1) == s.end() && last != -1) {
            s.insert({r+1,last});
            ms.insert(r+1-last);
        }
        /*cout << "set" << endl;
        for(auto x : s) cout << x.first << ' ' << x.second;
        cout << endl;
        cout << "multiset" << endl;
        for(auto x : ms) cout << x << ' ';
        cout << endl;*/
        if(ms.empty()) cout << "-1" << endl;
        else cout << i + *ms.begin() << endl;
    }
    
    return 0;
}