//Jakon; Two Pointers and Difference
#include <bits/stdc++.h>

using namespace std;

const int N = 200010;

int n, m, cnt[N], ans[N];

vector<int> v[N];

int main()

{

    cin >> n >> m;

    for(int i = 1; i <= n; i++)

    {

        int a, b;

        scanf("%d%d", &a, &b);

        v[a].push_back(i), v[b].push_back(i);

    }

    int sum = 0;

    for(int i = 1, j = 0; i <= m && j <= m; i++)

    {

        //以i为左边界，找到满足条件的最小右边界j

        while(sum < n && j < m) {

            ++ j;

            for(auto& idx : v[j]) {

                ++ cnt[idx];

                if(cnt[idx] == 1) ++ sum;

            }

        }

        //若区间[i, j]满足条件，则长度在[j-i+1, m-i+1]的答案都+1，即以i为左边界的所有可能情况都+1

        if(sum == n && j <= m) {

            ++ ans[j - i + 1], -- ans[m - i + 2];

        }

        //左边界i要右移一位，于是先把原本i的贡献去掉

        for(auto& idx : v[i]) {

            -- cnt[idx];

            if(cnt[idx] == 0) -- sum;

        }

    }

    for(int i = 1; i <= m; i++) cout << (ans[i] += ans[i - 1]) << " ";

    cout << endl;

    return 0;

}