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, a) for (int i = (int)0; i < (int)a; ++i)
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    int H, W;
    cin >> H >> W;

    map<int, int> candidates;
    map<int, int> moves;

    rep(i, W) {
        candidates[i] = i;
    }
    moves[0] = W;

    rep(i, H) {
        int A, B;
        cin >> A >> B;
        A--;

        int r = -1;
        while (true) {
            auto it = candidates.lower_bound(A);
            if (it == candidates.end() || it->first > B) break;

            r = max(r, it->second);
            int t = it->first - it->second;
            moves[t]--;
            if (moves[t] == 0) {
                moves.erase(t);
            }
            candidates.erase(it);
        }
        if (r != -1 && B != W) {
            moves[B - r]++;
            candidates[B] = r;
        }

        if (moves.size() == 0) {
            cout << -1 << "\n";
        } else {
            cout << moves.begin()->first + (i + 1) << "\n";
        }
    }
    cout << flush;

    return 0;
}