神犇有一个n个节点的图。因为神犇是神犇，所以在T时间内一些边会出现后消失。神犇要求出每一时间段内这个图是否是二分图。这么简单的问题神犇当然会做了，于是他想考考你。

输入数据的第一行是三个整数n,m,T。

第2行到第m+1行，每行4个整数u,v,start,end。第i+1行的四个整数表示第i条边连接u,v两个点，这条边在start时刻出现，在第end时刻消失。

输出包含T行。在第i行中，如果第i时间段内这个图是二分图，那么输出“Yes”，否则输出“No”，不含引号。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 100010
using namespace std;
struct data
{int x , y , t1 , t2;
}a[N << 1];
int n , fa[N << 2] , c[2][N << 2] , si[N << 2] , w[N << 2] , mp[N << 2] , rev[N << 2] , num[N] , sum , now;
inline char nc()
{static char buf[100000] , *p1 , *p2;return p1 == p2 && (p2 = (p1 = buf) + fread(buf , 1 , 100000 , stdin) , p1 == p2) ? EOF : *p1 ++ ;
}
inline int read()
{int ret = 0; char ch = nc();while(ch < '0' || ch > '9') ch = nc();while(ch >= '0' && ch <= '9') ret = (ret << 3) + (ret << 1) + ch - 48 , ch = nc();return ret;
}
bool cmp(data a , data b)
{return a.t1 < b.t1;
}
void pushup(int x)
{si[x] = si[c[0][x]] + si[c[1][x]] + 1;mp[x] = x;if(w[mp[c[0][x]]] < w[mp[x]]) mp[x] = mp[c[0][x]];if(w[mp[c[1][x]]] < w[mp[x]]) mp[x] = mp[c[1][x]];
}
void pushdown(int x)
{if(rev[x]){int l = c[0][x] , r = c[1][x];swap(c[0][l] , c[1][l]) , swap(c[0][r] , c[1][r]);rev[l] ^= 1 , rev[r] ^= 1 , rev[x] = 0;}
}
bool isroot(int x)
{return c[0][fa[x]] != x && c[1][fa[x]] != x;
}
void update(int x)
{if(!isroot(x)) update(fa[x]);pushdown(x);
}
void rotate(int x)
{int y = fa[x] , z = fa[y] , l = (c[1][y] == x) , r = l ^ 1;if(!isroot(y)) c[c[1][z] == y][z] = x;fa[x] = z , fa[y] = x , fa[c[r][x]] = y , c[l][y] = c[r][x] , c[r][x] = y;pushup(y) , pushup(x);
}
void splay(int x)
{update(x);while(!isroot(x)){int y = fa[x] , z = fa[y];if(!isroot(y)){if((c[0][y] == x) ^ (c[0][z] == y)) rotate(x);else rotate(y);}rotate(x);}
}
void access(int x)
{int t = 0;while(x) splay(x) , c[1][x] = t , pushup(x) , t = x , x = fa[x];
}
int find(int x)
{access(x) , splay(x);while(c[0][x]) pushdown(x) , x = c[0][x];return x;
}
void makeroot(int x)
{access(x) , splay(x);swap(c[0][x] , c[1][x]) , rev[x] ^= 1;
}
void link(int x , int y)
{makeroot(x) , fa[x] = y;
}
void cut(int x , int y)
{makeroot(x) , access(y) , splay(y) , fa[x] = c[0][y] = 0 , pushup(y);
}
void split(int x , int y)
{makeroot(x) , access(y) , splay(y);
}
void add(int p)
{int tx = a[p].x , ty = a[p].y , tmp , flag = 0;if(tx == ty && a[p].t2 > now) num[a[p].t2] ++ , sum ++ ;else{if(find(tx) != find(ty)) link(p + n , tx) , link(p + n , ty);else{split(tx , ty);if(!((si[ty] >> 1) & 1)) flag = 1;if(w[mp[ty]] >= a[p].t2) tmp = p;else tmp = mp[ty] - n , cut(tmp + n , a[tmp].x) , cut(tmp + n , a[tmp].y) , link(p + n , tx) , link(p + n , ty);if(flag && a[tmp].t2 > now) num[a[tmp].t2] ++ , sum ++ ;}}
}
int main()
{int m , t , i , p = 1;n = read() , m = read() , t = read();for(i = 1 ; i <= m ; i ++ ) a[i].x = read() , a[i].y = read() , a[i].t1 = read() , a[i].t2 = read();sort(a + 1 , a + m + 1 , cmp);for(i = 1 ; i <= n + m ; i ++ ) si[i] = 1 , mp[i] = i;for(i = 0 ; i <= n ; i ++ ) w[i] = 1 << 30;for(i = 1 ; i <= m ; i ++ ) w[i + n] = a[i].t2;for(i = 0 ; i < t ; i ++ ){now = i;while(p <= m && a[p].t1 <= i) add(p ++ );sum -= num[i];if(sum) puts("No");else puts("Yes");}return 0;
}