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>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
using namespace std;
using ll = long long ;
using P = pair<int,int> ;
using pll = pair<long long,long long>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e17;
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;

int main(){
    int h,w;
    cin >> h >> w;
    vector<int> a(h),b(h);
    rep(i,h){
        cin >> a[i] >> b[i];
        --a[i];--b[i];
    }

    map<int,int> mp;
    multiset<int> st;
    rep(i,w){
        mp[i]=i;
        st.insert(0);
    }
    vector<int> ans(h,-1);

    rep(i,h){
        auto it = mp.lower_bound(a[i]);

        int mr = -1;
        while(it != mp.end() && (it->first) <= b[i]+1){
            mr = max(mr,it->second);
            st.erase(st.find(it->first-it->second));
            it = mp.erase(it);
        }

        if(mr != -1 && b[i] < w-1){
            st.insert(b[i]+1-mr);
            mp[b[i]+1] = mr;
        }

        if((int)st.size() > 0){
            ans[i] = *st.begin() + i + 1;
        }
    }

    rep(i,h) cout << ans[i] << endl;
    
    return 0;
}