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 <iostream>
#include <vector>
#include <algorithm>
#include <climits>
#include <set>

typedef long long ll;

using namespace std;

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

    set<pair<int, int>> s;
    multiset<int> m;
    for (int i = 0; i < W; ++i) {
        s.insert({i, i});
        m.insert(0);
    }

    for (int i = 1; i <= H; ++i) {
        int A, B;
        cin >> A >> B;
        --A;
        --B;

        if (B == W - 1) {
            auto it = s.lower_bound({A, -1});
            for (; it != s.end() && it->first <= B;) {
                m.erase(m.find(it->first - it->second));
                s.erase(it++);
            }
        }
        else {
            int last_start_point = -1;
            auto it = s.lower_bound({A, -1});
            for (; it != s.end() && it->first <= B;) {
                last_start_point = it->second;
                m.erase(m.find(it->first - it->second));
                s.erase(it++);
            }
            if (last_start_point != - 1 && (it == s.end() || it->first != B + 1)) {
                m.insert(B + 1 - last_start_point);
                s.insert({B + 1, last_start_point});
            }
        }

        cout << (m.empty() ? -1 : (*m.begin() + i)) << endl;
    }
    return 0;
}