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;
    map<int, int> m;
    multiset<int> v;
    for(int i = 0; i < w; i++)
        m[i] = i;
    for(int i = 0; i < w; i++)
        v.insert(0);
    for(int i = 0; i < h; i++){
        int a, b;
        cin >> a >> b;
        a--;
        auto it = m.lower_bound(a);
        int mm = -1;
        while(it != m.end() && it->first <= b){
            mm = max(mm, it->second);
            int x = it->first - it->second;
            v.erase(v.find(x));
            m.erase(it++);
        }
        if(mm != -1 && b < w){
            v.insert(b - mm);
            m[b] = mm;
        }
        int ans = -1;
        if(v.size() > 0)
            ans = *v.begin() + (i + 1);
        cout << ans << endl;
    }
    return 0;
}