// ubsan: undefined
// accoders
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
int n,m;
struct Seg {
    int l, r,id;
    friend bool operator<(const Seg &a, const Seg &b) { return a.l < b.l; }
} a[N];
inline int find(int x) {
    int l = x, r = n, ans = n + 1;
    while (l <= r) {
        int mid = l + r >> 1;
        if (a[mid].l > a[x].r)
            r = mid - 1, ans = mid;
        else
            l = mid + 1;
    }
    return ans;
}
int Len[N];
inline bool cmp(int x, int y) { return Len[x] < Len[y]; }
int b[N], vis[N];
int ans, pos;
struct node{
	int a,b;
}Ans[N];
inline int check(int x) {
    int ans = 0;
    pos = x;
    memset(vis, 0, sizeof vis);
    for (int i = 1; i <= n; i++) {
        if (vis[i])
            continue;
        int j = max(Len[b[i]], pos);
        while (vis[j]) j++;
        if (j > n)
            break;
        ans++;
		vis[i]=vis[j]=ans;
		//cout<<i<<" "<<j<<" "<<ans<<endl;
        pos = j;
    }
    return ans;
}
inline int solve() {
    int l = 1, r = n, ans = 0,pos=0;
    while (r - l >= 5) {
        int mid = (r - l) / 3;
        int lmid = l + mid;
        int rmid = r - mid;
        if (check(lmid) > check(rmid))
            r = rmid - 1;
        else
            l = lmid + 1;
    }
    for (int i = l; i <= r; i++){
    	int val=check(i);
    	if(val>ans)ans=val,pos=i;
	}
    return pos;
}
int o[N],p[N];
int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i].l >> a[i].r,a[i].id=i;
    sort(a + 1, a + 1 + n);
    for (int i = 1; i <= n; i++) Len[i] = find(i), cout;
    for (int i = 1; i <= n; i++) b[i] = i;
    sort(b + 1, b + 1 + n, cmp);
    for(int i=1;i<=n;i++){
    	p[b[i]]=i;
	} 
    int pos=solve();
    int val=check(pos);
    if(val<m){
    	for(int i=1;i<=n;i++){
			if(vis[i]==0){
				if(val<m)val++;
				else continue;
				vis[i]=val;
			}
		
		} 
	}else{
		for(int i=1;i<=n;i++){
			if(vis[i]>m)vis[i]=0;
		}
	}
    for(int i=1;i<=n;i++){
    	o[a[i].id]=vis[p[i]];
	}
	for(int i=1;i<=n;i++)cout<<o[i]<<" ";
}
/*
对于一张有特殊性质的图求最大匹配
左端点排序
每个点都会有一段后缀可以匹配
把每个点可以最后匹配的后缀长度从大到小排一下，
挨个分配应该就好了
不然某个节点可能失配

好像还得三分一下能匹配点的范围，
5
9 13
3 7
9 12
7 10
3 7
*/