题目大意：给出M条管道，每条管道都有相应的下界和上界
现在问每条管道之间形成一个环，管道之间流量的最大值是多少

解题思路：两点之间的有向边的容量就是边的上界－下界
然后添加一个超级源点，连线那些流入的值大于流出的值的点，容量为流入值－流出值
再添加一个超级汇点，连线那些流入值小于流出值的点，容量为流出值－流入值
接着跑最大流，如果满流，表示可行
要求每条边的流量，只要找出每条边的流量，再加上每条边的下界即可

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXNODE = 210;
const int MAXEDGE = 100010;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge{
    int u,v,next,id;
    Type cap,flow;
    Edge() {}
    Edge(int u,int v,Type cap,Type flow,int next,int id) : u(u),v(v),cap(cap),flow(flow),next(next),id(id){}
};

struct Dinic{
    int n,m,s,t;
    Edge edges[MAXEDGE];
    int head[MAXNODE];
    int cur[MAXNODE];
    bool vis[MAXNODE];
    Type d[MAXNODE];
    vector<int> cut;

    void init(int n) {
        this->n = n;
        memset(head,-1,sizeof(head));
        m = 0;
    }

    void AddEdge(int u,int id) {
        edges[m] = Edge(u,cap,0,head[u],id);
        head[u] = m++;
        edges[m] = Edge(v,u,head[v],id);
        head[v] = m++;
    } 

    bool BFS() {
        memset(vis,sizeof(vis));
        queue<int> Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = 1;

        while (!Q.empty()) {
            int u = Q.front(); Q.pop();
            for (int i = head[u]; ~i; i = edges[i].next) {
                Edge &e = edges[i];
                if (!vis[e.v] && e.cap > e.flow) {
                    vis[e.v] = true;
                    d[e.v] = d[u] + 1;
                    Q.push(e.v);
                }
            }
        }
        return vis[t];
    }

    Type DFS(int u,Type a) {
        if (u == t || a == 0) return a;

        Type flow = 0,f;
        for (int &i = cur[u]; i != -1; i = edges[i].next) {
            Edge &e = edges[i];
            if (d[u] + 1 == d[e.v] && (f = DFS(e.v,min(a,e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[i ^ 1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }

    Type Maxflow(int s,int t) {
        this->s = s; this->t = t;
        Type flow = 0;
        while (BFS()) {
            for (int i = 0; i < n; i++)
                cur[i] = head[i];
            flow += DFS(s,INF);
        }
        return flow;
    }

    void Mincut() {
        cut.clear();
        for (int i = 0; i < m; i += 2) {
            if (vis[edges[i].u] && !vis[edges[i].v]) 
                cut.push_back(i);
        }
    }

    bool judge() {
        for (int i = head[s]; ~i; i = edges[i].next) 
            if (edges[i].cap - edges[i].flow != 0)
                return false;
        return true;
    }
}dinic;

int n,m;
int d[MAXNODE],ans[MAXEDGE];

void solve() {
    memset(d,sizeof(d));
    int source = n + 1,sink = n + 2;
    dinic.init(sink + 1);

    int u,up,down;
    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d%d",&u,&v,&down,&up);
        dinic.AddEdge(u,up - down,i);
        ans[i] = down;
        d[u] -= down;
        d[v] += down;
    }
    int tot = dinic.m;
    for (int i = 1; i <= n; i++) {
        if (d[i] > 0) dinic.AddEdge(source,i,d[i],0);
        if (d[i] < 0) dinic.AddEdge(i,sink,-d[i],0);
    }
    dinic.Maxflow(source,sink);
    if (!dinic.judge()) printf("NO\n");
    else {
        for (int i = 0; i < tot; i += 2) 
            ans[dinic.edges[i].id] += dinic.edges[i].flow;
        printf("YES\n");
        for (int i = 1; i <= m; i++)
            printf("%d\n",ans[i]);
    }
}

int main() {
    while (~scanf("%d%d",&n,&m)) solve();
    return 0;
}