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;
template <class T> using vec = vector<T>;
template <class T> using vvec = vector<vec<T>>;

constexpr int inf = 1e9;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int H,W;
    cin >> H >> W;
    vec<int> A(H),B(H);
    set<pair<int,int>> s;
    multiset<int> ms;
    for(int i=0;i<H;i++){
        cin >> A[i] >> B[i];
        A[i]--, B[i]--;
    }
    for(int i=0;i<W;i++){
        ms.insert(0);
        s.insert({i,-i});
    }
    for(int k=0;k<H;k++){
        auto it = s.upper_bound({A[k],-inf});
        while(it!=s.end()){
            auto [en,st] = *it;
            if(en>B[k]) break;
            int ne = (B[k]!=W-1? B[k]+1:inf);
            s.insert({ne,st});
            ms.insert(ne+st);
            it = s.erase(it);
            ms.erase(ms.find(en+st));
        }
        if(B[k]!=W-1){
            it = s.upper_bound({B[k]+1,-inf});
            it++;
            while(it!=s.end()){
                auto [en,st] = *it;
                if(en>B[k]+1) break;
                it = s.erase(it);
                ms.erase(ms.find(en+st));
            }
        }
        int res = *ms.begin()+k+1;
        cout << (res>H+W? -1:res) << "\n";
    }
}