题意：一个人要给女孩子们拍照，一共 n 天，m 个女孩子，每天他至多拍 d[i] 张照片，每个女孩子总共要被至少拍 g[i] 次。在第 i 天，可以拍 c[i] 个女孩子，c[i] 个女孩子中每个女孩子在当天被拍的次数是 [li,ri]，求最多可以拍多少张照片，以及每天每个可以拍的女孩子被拍了多少张照片。

析：一个很明显网络流，很容易建图，建立一个源点 s 和汇点 t，然后 s 向每一天连一条边，容量上界上 d[i]，每个女孩子向汇点连一边，下界是 g[i]， 上界是无穷大，然后对于每一天拍的从每一天那个结点向那个女孩子连一条容量上界是 ri，下界是 li，的边，然后就建好图了，就是求最大流，但是这个不是普通的网络流，我们要把它进行转换，首先先根据无源无汇的，建立，因为是有源汇的，所以连一条边 t -> s，INF，然后求一次 超级源点汇点（注意不是 s t） S T 最大流，如果满流就是有可行流，然后再求 s->t 的最大流，结果就是答案。然后再输出解，输出解的时候就是流再加上下界。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1500 + 50;
const int maxm = 1e6 + 10;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }

struct Edge{
  int from, to, cap, flow;
};

struct Dinic{
  int n, m, s, t;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool vis[maxn];
  int d[maxn];
  int cur[maxn];
  
  void init(int n){
    FOR(i, n, 0)  G[i].cl;
    edges.cl;
  }
  
  void addEdge(int from, int to, int cap){
    edges.pb((Edge){from, to, cap, 0});
    edges.pb((Edge){to, from, 0, 0});
    m = edges.sz;
    G[from].pb(m - 2);
    G[to].pb(m - 1); 
  }
  
  bool bfs(){
    ms(vis, 0);  d[s] = 0;  vis[s] = 1;
    queue<int> q;  q.push(s);
    
    while(!q.empty()){
      int u = q.front();  q.pop();
      for(int i = 0; i < G[u].sz; ++i){
        Edge &e = edges[G[u][i]];
        if(!vis[e.to] && e.cap > e.flow){
          vis[e.to] = 1;
          d[e.to] = d[u] + 1;
          q.push(e.to);
        }
      }
    }
    return vis[t];
  }
  
  int dfs(int u, int a){
    if(u == t || a == 0)  return a;
    int flow = 0, f;
    for(int &i = cur[u]; i < G[u].sz; ++i){
      Edge &e = edges[G[u][i]];
      if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
        e.flow += f;
        edges[G[u][i]^1].flow -= f;
        flow += f;
        a -= f;
        if(a == 0)  break;
      }
    }
    return flow;
  }
  
  int maxFlow(int s, int t){
    this->s = s; this->t = t;
    int flow = 0;
    while(bfs()){ ms(cur, 0);  flow += dfs(s, INF); }
    return flow;
  }
};
Dinic dinic;

int down[100000], in[100000], d[maxn], g[maxn];

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    int s = 0, t = n + m + 1;
    dinic.init(t + 10);
    ms(in, 0);
    for(int i = 1; i <= m; ++i)
      scanf("%d", g + i);
    
    int cnt = 0;
    for(int i = 1; i <= n; ++i){
      int c;  scanf("%d %d", &c, d + i);
      int x, l, r;
      while(c--){
        scanf("%d %d %d", &x, &l, &r);
        down[cnt++] = l;
        dinic.addEdge(i, x + n + 1, r - l);
        in[i] -= l;
        in[x+n+1] += l;
      }
    }
    for(int i = 1; i <= n; ++i)
      dinic.addEdge(s, i, d[i]);
    for(int i = 1; i <= m; ++i){
      dinic.addEdge(i + n, t, INF - g[i]);
      in[i+n] -= g[i];
      in[t] += g[i];
    }
    int S = t + 1, T = t + 2;
    int ans = 0;
    for(int i = 0; i <= t; ++i)
      if(in[i] > 0)  dinic.addEdge(S, i, in[i]), ans += in[i];
      else if(in[i] < 0)  dinic.addEdge(i, T, -in[i]);
    dinic.addEdge(t, s, INF);
    if(ans != dinic.maxFlow(S, T)){ printf("-1\n\n");  continue; }
    printf("%d\n", dinic.maxFlow(s, t));
    for(int i = 0; i < cnt; ++i)  printf("%d\n", dinic.edges[i<<1].flow + down[i]);
    printf("\n");
  }
  return 0;
}