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, s, n) for (int i = s; i < (int)(n); i++)
#define ALL(a) a.begin(), a.end()
#define MOD 1000000007
using namespace std;
using ll = long long;

int main() {
    int H, W; cin >> H >> W;

    map<int, int> mp;
    multiset<int> ms;

    REP(i, 0, W) {
        mp[i] = i;
        ms.insert(0);
    }

    REP(i, 0, H) {
        int a, b; cin >> a >> b;
        a--;

        int right = -1;
        auto itr = mp.lower_bound(a);
        while (itr != mp.end() && itr->first <= b) {
            right = itr->second;
            ms.erase(ms.find(itr->first - itr->second));
            mp.erase(itr++);
        }

        if (right != -1 && b < W) {
            ms.insert(b - right);
            mp[b] = right;
        }

        if (ms.size()) cout << *ms.begin() + i + 1 << endl;
        else cout << -1 << endl;
    }
    return 0;
}