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;

#define REP(i,p,n) for(ll i=p;i<ll(n);++i)
#define rep(i,n) REP(i,0,n)

using ll = long long;
using pll = pair<ll, ll>;

const ll INF = 1LL << 30;

int main()
{
    ll H, W;
    cin >> H >> W;

    set<pll> nodes;
    multiset<ll> costs;

    rep (i, W)
    {
        nodes.emplace(i, i);
        costs.emplace(0);
    }

    rep (y, H)
    {
        ll a, b;
        cin >> a >> b, --a, --b;

        ll newSrc = -1;
        const auto newDst = b + 1;

        // ひとつだけ残して削除する (TODO: 範囲削除)
        {
            auto it = nodes.lower_bound(pll(a, 0));
            while (it != nodes.end())
            {
                const auto src = it->second, dst = it->first;
                if (b < dst) { break; }

                // nodesとcostsは同期してるので必ずdst-srcな値があるはず
                costs.erase(costs.find(dst - src));
                it = nodes.erase(it);

                newSrc = max(newSrc, src);
            }
        }


        if (newSrc != -1 && newDst != W)
        {
            nodes.emplace(newDst, newSrc);
            costs.emplace(newDst - newSrc);
        }

        const auto cost = costs.begin() != costs.end() ? *costs.begin() : W;
        cout << (cost != W ? cost + y + 1 : -1) << endl;
    }

    return 0;
}