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;
const int N = 2e5 + 10;
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    map <int, int> l;
    multiset <int> ans;
    for (int i = 1; i <= m; ++i)
    {
        l[i] = i;
        ans.emplace(0);
    }
    /*
        for (auto& i : l)
        {
            cout << i.second << '\t' << i.first << endl;
        }
    */
    for (int i = 1; i <= n; ++i)
    {
        int a, b;
        cin >> a >> b;
        ++b;
        auto it = l.lower_bound(a);
        int r = -1;
        for (; it != l.end() && it->first <= b; it = l.erase(it))
        {
            ans.erase(ans.find(it->first - it->second));
            r = max(r, it->second);
        }
        if (r != -1 && b <= m)
        {
            l[b] = r;
            ans.emplace(b - r);
        }
        cout << (ans.empty() ? -1 : *ans.begin() + i) << endl;
        /*
        cout << r << endl;
        for (auto& i : l)
        {
            cout << i.second << '\t' << i.first << endl;
        }
        cout << endl;
        */
    }
}