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 ll = long long int;
using P = std::pair<int, int>;

#define rep(i, n) for (int i = 1; i <= (n); ++i)
#define rag(con) std::begin(con), std::end(con)

int main()
{

    int h, w;

    std::cin >> h >> w;

    std::multiset<int> set;
    rep(i, w)
        set.insert(0);
    std::map<int, int> mp; //終点；始点

    rep(i, w)
        mp[i] = i;

    rep(i, h)
    {
        int a, b;

        std::cin >> a >> b;

        auto bg = mp.lower_bound(a), ed = mp.upper_bound(b);

        int mx = 0;

        for (auto it = bg; it != ed; ++it)
        {
            mx = it->second;
            set.erase(set.find(it->first - it->second));
        }

        //std::cerr << "D " << mx << std::endl;

        if (mx != 0)
        {

            mp.erase(bg, ed);

            if (b < w)
            {

                if (mp[b + 1] < mx)
                {
                    mp[b + 1] = mx;

                    set.insert(b + 1 - mx);
                }
            }
        }

        if (set.empty())
            std::cout << -1 << std::endl;
        else
            std::cout << i + *(set.begin()) << std::endl;

        //std::cerr << "DEBUG" << std::endl;
        /*for (auto x : mp)
            std::cerr << x.first << ' ' << x.second << std::endl;*/
    }

    return 0;
}