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)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
    // Happy Candle、それぞれの地点からスタートしたものを保持
    // (現在地、スタート)のpairのset、右に移動した距離のmultiset
    int H, W;
    cin >> H >> W;
    map<int, int> pieces;
    multiset<int> rmove;
    rep(i, W) {
        pieces[i] = i;
        rmove.insert(0);
    }
    rep(i, H) {
        int a, b;
        cin >> a >> b;
        a--;
        auto it = pieces.lower_bound(a);
        int limit = -1;
        while (it != pieces.end() && it->first <= b) {
            limit = max(limit, it->second);
            int move = it->first - it->second;
            pieces.erase(it++);
            rmove.erase(rmove.find(move));
        }
        if (limit != -1 && b < W) {
            pieces[b] = limit;
            rmove.insert(b-limit);
        }
        if (!rmove.empty()) {
            cout << i + 1 + *rmove.begin() << '\n';
        } else cout << -1 << '\n';
    }
    return 0;
}

// 09