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 LL long long
#define ULL unsigned long long
#define UI unsigned int
#define mem(i, j) memset(i, j, sizeof(i))
#define rep(i, j, k) for(int i = j; i <= k; i++)
#define dep(i, j, k) for(int i = k; i >= j; i--)
#define pb push_back
#define make make_pair
#define INF 0x3f3f3f3f
#define inf LLONG_MAX
#define PI acos(-1)
#define fir first
#define sec second
#define lb(x) ((x) & (-(x)))
#define dbg(x) cout<<#x<<" = "<<x<<endl;
using namespace std;

const int N = 1e6 + 5;

int n, m;

int cnt[N], L[N], R[N];

map < int, int > mp;

void solve() {

    scanf("%d %d", &n, &m);

    rep(i, 1, n) scanf("%d %d", &L[i], &R[i]);

    rep(i, 0, m - 1) mp[i] = i;

    cnt[0] = m;

    int ans = 0;

    rep(i, 1, n) {

        auto it = mp.lower_bound(L[i] - 1);

        int r = -1;

        while(it != mp.end() and it->fir <= R[i]) {

            cnt[it->fir - it->sec]--;

            r = max(r, it->sec);

            it = mp.erase(it);

        }

        if(r >= 0 && R[i] < m) {

            mp[R[i]] = r;

            cnt[R[i] - r]++;

        }

        while(ans < m && cnt[ans] == 0) ans++;

        if(ans == m) puts("-1");

        else printf("%d\n", i + ans);

    }

}


int main() {

//    int _; scanf("%d", &_);
//    while(_--) solve();

    solve();

    return 0;
}