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, e) for(int (i) = 0; (i) < (e); ++(i))
#define FOR(i, b, e) for(int (i) = (b); (i) < (e); ++(i))
#define ALL(c) (c).begin(), (c).end()
#define PRINT(x) cout << (x) << "\n"
using namespace std;
using ll = long long; using pint = pair<int, int>; using pll = pair<ll, ll>;
const long long MOD = 1000000007;

ll H, W, A[200010], B[200010];

signed main(){
    cin >> H >> W;
    REP(i, H){
        cin >> A[i] >> B[i];
        A[i]--; B[i]--;
    }
    set<pll> st;
    multiset<ll> ms;
    REP(i, W){
        st.insert(pll(i, i));
        ms.insert(0);
    }
    REP(i, H){
        auto l = st.lower_bound(pll(A[i], -1));
        auto r = st.lower_bound(pll(B[i] + 1, -1));
        ll h = LONG_LONG_MIN;
        while(l != r){
            h = max(h, l->second);
            auto itr = ms.find(l->first - l->second);
            if(itr != ms.end()) ms.erase(itr);
            l = st.erase(l);
        }
        if(h >= 0 && B[i] < W - 1){
            st.insert(pll(B[i] + 1, h));
            ms.insert(B[i] + 1 - h);
        }
        if(ms.empty()) PRINT("-1");
        else PRINT(i + 1 + *ms.begin());
    }
    return 0;
}