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 <map>
#include <set>
#include <vector>
using namespace std;

void solve() {
    int H, W;
    cin >> H >> W;

    multiset<int> steps;
    map<int, int> end2start;
    for (int c = 0; c < W; c++) {
        end2start[c] = c;
        steps.insert(0);
    }

    for (int r = 1; r <= H; r++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;

        // Erase pairs (e, s) that a <= e <= b + 1
        auto lb = end2start.lower_bound(a);
        auto ub = end2start.upper_bound(b + 1);
        auto max_start = -1;
        for (auto it = lb; it != ub; ++it) {
            int end = it->first;
            int start = it->second;
            steps.erase(steps.find(end - start));
            max_start = max(max_start, start);
        }
        end2start.erase(lb, ub);

        // Update pair (b + 1, s)
        if (b != W - 1 and lb != ub) {
            end2start[b + 1] = max_start;
            steps.insert(b + 1 - end2start[b + 1]);
        }

        if (steps.size() > 0) {
            cout << *steps.begin() + r << endl;
        } else {
            cout << "-1" << endl;
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
    return 0;
}