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 int long long
#define db(a, ...) cerr<<#a<<" = "<<a<<'\n'
#define db2(a, b) cerr<<#a<<" = "<<a<<", ", db(b)
#define db3(a, b, c) cerr<<#a<<" = "<<a<<", ", db2(b, c)
#define db4(a, b, c, d) cerr<<#a<<" = "<<a<<", ", db3(b, c, d)

using namespace std;

typedef pair<int, int> ii;

const int INF = 1e9, MAX = 2e5 + 5;

int n, k;
int d[MAX];

int32_t main() {
#ifdef ACM
    freopen("D.in", "r", stdin);
#endif
    cin >> n >> k;
    multiset<int> S1;
    multiset<ii> S2;
    for (int i = 1; i <= k; i++) {
        S1.insert(0);
        S2.insert({i, 0});
    }
    int l, r;
    for (int i = 1; i <= n; i++) {
        cin >> l >> r;
        int minD = INF;
        vector<ii> V;
        for (auto it = S2.lower_bound({l, 0}); it != S2.end() && it->first <= r; it++) {
            int newD = it->second + r - it->first + 1;
            minD = min(minD, newD);
            V.emplace_back(it->second, it->first);
        }
        for (ii it:V) {
            S1.erase(S1.find(it.first));
            S2.erase(S2.find({it.second, it.first}));
        }
        if (r < k && minD < INF) {
            d[r + 1] = minD;
            S1.insert(minD);
            S2.insert({r + 1, d[r + 1]});
        }
        if (S1.size() == 0)
            cout << -1 << '\n';
        else
            cout << *S1.begin() + i << '\n';
    }
}