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;
using ll = long long;

const int INF = 1<<30;

template<class T> inline bool chmax(T &a, T b)
{
    if(a < b)
    {
        a = b;
        return true;
    }
    return false;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int H, W;
    cin >> H >> W;
    vector<int> A(H), B(H);
    for(int i=0; i<H; i++)
    {
        cin >> A[i] >> B[i];
        A[i]--;
        B[i]--;
    }

    map<int, int> path;
    multiset<int> dist;
    for(int i=0; i<W; i++)
    {
        path[i] = i;
        dist.insert(0);
    }

    for(int i=0; i<H; i++)
    {
        int idx = -1;
        for(auto itr = path.lower_bound(A[i]); itr != path.end() && itr->first <= B[i] + 1; itr = path.erase(itr))
        {
            chmax(idx, itr->second);
            auto ditr = dist.find(itr->first - itr->second);
            dist.erase(ditr);
        }
        if(idx != -1 && B[i] + 1 < W)
        {
            path[B[i] + 1] = idx;
            dist.insert(B[i] + 1 - idx);
        }
        if(dist.size() == 0)
        {
            cout << "-1\n";
        }
        else
        {
            cout << *dist.begin() + i + 1 << '\n';
        }
    }

    return 0;
}