/*
思维难度几乎没有， 就是线段树分治check二分图
判断是否为二分图可以通过维护lct看看是否链接出奇环
然后发现不用lct， 并查集维护奇偶性即可

但是复杂度明明一样哈

*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define f1 first
#define f2 second
#define ll long long
#define mmp make_pair
#define lson l, mid, now << 1
#define rson mid + 1, r, now << 1 | 1
#define ls now << 1
#define rs now << 1 | 1
#define M 100010
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
vector<pair<int, int> > seg[M << 2];

int n, m, t, father[M], sz[M], ans[M], dis[M];
int find(int x) {
    return father[x] == x ? x : find(father[x]);
}

int getdis(int x) {
    return father[x] == x ? 0 : getdis(father[x]) ^ dis[x];
}
void modify(int l, int r, int now, int ln, int rn, pair<int, int> v) {
    if(l > rn || r < ln) return;
    if(l >= ln && r <= rn) {
        seg[now].push_back(v);
        return;
    }
    int mid = (l + r) >> 1;
    modify(lson, ln, rn, v);
    modify(rson, ln, rn, v);
}

void work(int l, int r, int now) {
    int mid = (l + r) >> 1, f = 0;
    vector<pair<int, int> > tmp;
    for(int i = 0; i < seg[now].size(); i++) {
        int vi = seg[now][i].f1, vj = seg[now][i].f2;
        int a = find(vi), b = find(vj);
        if(a == b) {
            if((getdis(vi) ^ getdis(vj)) == 0) {
                f = 1;
                break;
            }
        } else {
            if(sz[a] > sz[b]) swap(a, b), swap(vi, vj);
            sz[b] += sz[a];
            dis[a] = dis[vi] ^ dis[vj] ^ 1;
            father[a] = b;
            tmp.push_back(mmp(a, b));
        }
    }
    if(!f) {
        if(l == r) ans[l] = 1;
        else work(lson), work(rson);
    }
    for(int i = tmp.size() - 1; i >= 0; i--) {
        int a = tmp[i].first, b = tmp[i].second;
        father[a] = a;
        dis[a] = 0;
        sz[b] -= sz[a];
    }
}
int main() {
    n = read(), m = read(), t = read();
    for(int i = 1; i <= n; i++) father[i] = i, sz[i] = 1;
    for(int i = 1; i <= m; i++) {
        int vi = read(), vj = read(), be = read(), ed = read();
        modify(1, t, 1, be + 1, ed, mmp(vi, vj));
    }
    work(1, t, 1);
    for(int i = 1; i <= t; i++) puts(ans[i] ? "Yes" : "No");
    return 0;
}
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