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 <algorithm>
#include <iostream>
#include <map>
#include <set>

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

    std::map<int, int> M;
    std::multiset<int> S;
    for (int i = 0; i < W; i++) {
        M[i] = i;
        S.insert(0);
    }

    for (int i = 0; i < H; i++) {
        int A, B;
        std::cin >> A >> B;
        A--;

        auto cur = M.lower_bound(A);
        int ma = -1;
        while (cur != M.end() and cur->first <= B) {
            ma = std::max(ma, cur->second);

            const int x = cur->first - cur->second;
            S.erase(S.find(x));
            M.erase(cur++);
        }

        if (ma != -1 and B < W) {
            S.insert(B - ma);
            M[B] = ma;
        }

        if (S.empty()) {
            std::cout << -1 << '\n';
        } else {
            std::cout << *S.begin() + i + 1 << '\n';
        }
    }
}