# Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

# Description

In the field of computer science, forest is important and deeply researched , it is a model for many data structures . Now it’s your job here to calculate the depth and width of given forests.

Precisely, a forest here is a directed graph with neither loop nor two edges pointing to the same node. Nodes with no edge pointing to are roots, we define that roots are at level 0 . If there’s an edge points from node A to node B , then node B is called a child of node A , and we define that B is at level (k+1) if and only if A is at level k .

We define the depth of a forest is the maximum level number of all the nodes , the width of a forest is the maximum number of nodes at the same level.

# Input

There’re several test cases. For each case, in the first line there are two integer numbers n and m (1≤n≤100, 0≤m≤100, m≤n*n) indicating the number of nodes and edges respectively , then m lines followed , for each line of these m lines there are two integer numbers a and b (1≤a,b≤n)indicating there’s an edge pointing from a to b. Nodes are represented by numbers between 1 and n .n=0 indicates end of input.

# Output

For each case output one line of answer , if it’s not a forest , i.e. there’s at least one loop or two edges pointing to the same node, output “INVALID”(without quotation mark), otherwise output the depth and width of the forest, separated by a white space.

# Sample Input

1 0
1 1
1 1
3 1
1 3
2 2
1 2
2 1
0 88


# Sample Output

0 1
INVALID
1 2
INVALID


# Problem Source

ZSUACM Team Member

 1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstring>
5 #include <queue>
6 #define rep(i,l,r) for(int i = l; i <= r; i++)
7 #define clr(x,y) memset(x,y,sizeof(x))
8 #define travel(x) for(Edge *p = last[x]; p; p = p -> pre)
9 using namespace std;
10 const int maxn = 110;
12     int ans = 0, f = 1; char c = getchar();
13     for(; !isdigit(c); c = getchar()) if (c == '-') f = -1;
14     for(; isdigit(c); c = getchar()) ans = ans * 10 + c - '0';
15     return ans * f;
16 }
17 struct Edge{
18     Edge *pre; int to;
19 }edge[10010];
20 Edge *last[maxn], *pt;
21 int n, m, inv[maxn], depth[maxn], width[maxn];
22 bool valid, vis[maxn];
23 queue <int> q;
24 inline void addedge(int x,int y){
25     pt->pre = last[x]; pt->to = y; last[x] = pt++;
26 }
27 void bfs(){
28     while (!q.empty()) q.pop();
29     rep(i,1,n) if (!inv[i]){
30         q.push(i); depth[i] = 0; width[0]++; vis[i] = 1;
31     }
32     if (q.empty()){
33         valid = 0; return;
34     }
35     while (!q.empty()){
36         int now = q.front(); q.pop();
37         travel(now){
38             if (vis[p->to]){
39                 valid = 0; return;
40             }
41             vis[p->to] = 1;
42             depth[p->to] = depth[now] + 1;
43             width[depth[p->to]]++;
44             q.push(p->to);
45         }
46     }
47     rep(i,1,n) if (!vis[i]) valid = 0;
48 }
49 void work(){
51     pt = edge; clr(last,0); clr(inv,0); clr(width,0); clr(depth,0); clr(vis,0); valid = 1;
52     rep(i,1,m){
55         inv[b]++;
56         if (inv[b] > 1) valid = 0;
57     }
58     if (!valid){
59         printf("INVALID\n");
60         return;
61     }
62     bfs();
63     if (!valid){
64         printf("INVALID\n");
65         return;
66     }
67     int maxdepth = 0; int maxwidth = 0;
68     rep(i,1,n){
69         maxdepth = max(maxdepth,depth[i]);
70         maxwidth = max(maxwidth,width[depth[i]]);
71     }
72     printf("%d %d\n",maxdepth,maxwidth);
73 }
74 int main(){
80 }