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,class U> inline bool chmin(T&x,U y){if(x>y){x=y;return true;}return false;}
template<class T,class U> inline bool chmax(T&x,U y){if(x<y){x=y;return true;}return false;}
#define fr(i,n) for(int i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)

void solve(){
    int h,w,a,b;
    cin>>h>>w;
    set<pair<int,int>> st;
    multiset<int> d;
    fr(i,w){
        st.emplace(i,i);
        d.emplace(0);
    }
    Fr(i,h){
        cin>>a>>b;--a;;
        auto it=st.lower_bound(make_pair(a,0));
        int l{-1};
        for(;it!=st.end()&&it->first<=b;){
            l=it->second;
            if(it->first==w) d.erase(INT_MAX);
            else d.erase(d.find(it->first-it->second));
            it=st.erase(it);
        }
        if(l>=0){
            st.emplace(b,l);
            d.emplace(b==w?INT_MAX:b-l);
        }
        cout<<(*d.begin()==INT_MAX?-1:*d.begin()+i)<<'\n';
    }
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    solve();
}