题意：带增删边的查询二分图
题解：因为二分图肯定带奇环，lct维护，每次要加入一条边之前判断会不会构成环，如果会就把最先会删除的边删掉，然后如果是奇环就打个标记，然后把奇环数++，删除的时候，把标记删除，然后奇环数量--，需要注意的是可能有自环，还有一点就是我们先把边拆成点了，判断奇环的时候不能直接看是奇数还是偶数，我们先/2再看是不是奇数，如果/2是偶数就是有奇环
/**************************************************************
Problem: 4025
User: walfy
Language: C++
Result: Accepted
Time:11120 ms
Memory:45776 kb
****************************************************************/
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}
using namespace std;
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=500000+10,maxn=100000+10,inf=0x3f3f3f3f;
struct LCT{
int fa[N],ch[N][2],rev[N],sz[N],q[N],odd[N],ti[N],id[N],mi[N];
void init()
{
memset(ch,0,sizeof ch);
memset(fa,0,sizeof fa);
memset(sz,0,sizeof sz);
memset(rev,0,sizeof rev);
}
inline bool isroot(int x){return ch[fa[x]][0]!=x&&ch[fa[x]][1]!=x;}
inline void pushup(int x)
{
sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+1;
mi[x]=ti[x];id[x]=x;
if(ch[x][0]&&mi[x]>mi[ch[x][0]])
{
mi[x]=mi[ch[x][0]];
id[x]=id[ch[x][0]];
}
if(ch[x][1]&&mi[x]>mi[ch[x][1]])
{
mi[x]=mi[ch[x][1]];
id[x]=id[ch[x][1]];
}
}
inline void pushdown(int x)
{
if(rev[x])
{
rev[x]=0;swap(ch[x][0],ch[x][1]);
rev[ch[x][0]]^=1,rev[ch[x][1]]^=1;
}
}
inline void Rotate(int x)
{
int y=fa[x],z=fa[y],l,r;
if(ch[y][0]==x)l=0,r=l^1;
else l=1,r=l^1;
if(!isroot(y))
{
if(ch[z][0]==y)ch[z][0]=x;
else ch[z][1]=x;
}
fa[x]=z;fa[y]=x;fa[ch[x][r]]=y;
ch[y][l]=ch[x][r];ch[x][r]=y;
pushup(y);
}
inline void splay(int x)
{
int top=1;q[top]=x;
for(int i=x;!isroot(i);i=fa[i])q[++top]=fa[i];
for(int i=top;i;i--)pushdown(q[i]);
while(!isroot(x))
{
int y=fa[x],z=fa[y];
if(!isroot(y))
{
if((ch[y][0]==x)^(ch[z][0]==y))Rotate(x);
else Rotate(y);
}
Rotate(x);
}
pushup(x);
}
inline void access(int x){for(int y=0;x;y=x,x=fa[x])splay(x),ch[x][1]=y,pushup(x);}
inline void makeroot(int x){access(x),splay(x),rev[x]^=1;}
inline int findroot(int x){access(x),splay(x);while(ch[x][0])x=ch[x][0];return x;}
inline void split(int x,int y){makeroot(x),access(y),splay(y);}
inline void cut(int x,int y){split(x,y);if(ch[y][0]==x)ch[y][0]=0,fa[x]=0;}
inline void link(int x,int y){makeroot(x),fa[x]=y,splay(x);}
}lct;
struct node{int u,v,st,en,id;}p[N];
vi s[N],e[N];
int main()
{
// fin;
int n,m,T;
scanf("%d%d%d",&n,&m,&T);
for(int i=1;i<=n;i++)lct.ti[i]=lct.mi[i]=inf,lct.id[i]=i,lct.sz[i]=1;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d%d",&p[i].u,&p[i].v,&p[i].st,&p[i].en);
p[i].id=i+n;
s[p[i].st].pb(i),e[p[i].en].pb(i);
lct.ti[i+n]=lct.mi[i+n]=p[i].en;lct.id[i+n]=i+n;lct.sz[i+n]=1;
}
int o=0,f=0;
for(int i=0;i<=T;i++)
{
if(i)puts(o?"No":"Yes");
// printf("%d %d
",i,o);
for(int j=0;j<s[i].size();j++)
{
node te=p[s[i][j]];
if(te.u==te.v)
{
lct.odd[te.id]=1,o++;
continue;
}
int x=lct.findroot(te.u),y=lct.findroot(te.v);
if(x!=y)lct.link(te.u,te.id),lct.link(te.v,te.id);
else
{
lct.split(te.u,te.v);
int k=lct.id[te.v]-n;
if(lct.mi[te.v]>te.en)
{
if((lct.sz[te.v]/2)%2==0)lct.odd[te.id]=1,o++;
continue;
}
if((lct.sz[te.v]/2)%2==0)lct.odd[k+n]=1,o++;
lct.cut(p[k].u,p[k].id),lct.cut(p[k].v,p[k].id);
lct.link(te.u,te.id),lct.link(te.v,te.id);
}
}
for(int j=0;j<e[i].size();j++)
{
node te=p[e[i][j]];
if(te.u==te.v)
{
o-=lct.odd[te.id];
lct.odd[te.id]=0;
continue;
}
int x=lct.findroot(te.u),y=lct.findroot(te.v);
if(x!=y)o-=lct.odd[te.id],lct.odd[te.id]=0;
else
{
lct.split(te.u,te.v);
// if((lct.sz[te.v]/2)%2==0)
o-=lct.odd[te.id],lct.odd[te.id]=0;
int k=lct.id[te.v]-n;
lct.cut(p[k].u,p[k].id),lct.cut(p[k].v,p[k].id);
}
}
}
return 0;
}
/********************
3 3 5
1 2 0 5
2 3 0 4
1 1 1 3
********************/