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>
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.emplace(i + 1, i + 1);
        m.insert(0);
    }
    for(int i = 0; i < h; i++) {
        int a, b;
        cin >> a >> b;
        int t = 0;
        auto it = s.lower_bound({a, 0}), jt = it;
        for(; jt != s.end(); ++jt) {
            auto [y, x] = *jt;
            if(y > b)
                break;
            t = x;
            m.erase(m.find(y - x));
        }
        s.erase(it, jt);
        if(t && b < w)
            s.emplace(b + 1, t), m.insert(b + 1 - t);
        cout << (m.empty() ? -1 : *m.begin() + i + 1) << '\n';
    }
    return 0;
}