codeforces 427c(强连通分量)

C. Checkposts

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Your city has n junctions. There are m one-way roads between the junctions. As a mayor of the city, you have to ensure the security of all the junctions.

To ensure the security, you have to build some police checkposts. Checkposts can only be built in a junction. A checkpost at junction ican protect junction j if either i = j or the police patrol car can go to j from i and then come back to i.

Building checkposts costs some money. As some areas of the city are more expensive than others, building checkpost at some junctions might cost more money than other junctions.

You have to determine the minimum possible money needed to ensure the security of all the junctions. Also you have to find the number of ways to ensure the security in minimum price and in addition in minimum number of checkposts. Two ways are different if any of the junctions contains a checkpost in one of them and do not contain in the other.

Input

In the first line, you will be given an integer n, number of junctions (1 ≤ n ≤ 105). In the next line, n space-separated integers will be given. The ith integer is the cost of building checkpost at the ith junction (costs will be non-negative and will not exceed 109).

The next line will contain an integer m (0 ≤ m ≤ 3·105). And each of the next m lines contains two integers ui andvi (1 ≤ ui, vi ≤ n; u ≠ v). A pair ui, vi means, that there is a one-way road which goes from ui to vi. There will not be more than one road between two nodes in the same direction.

Output

Print two integers separated by spaces. The first one is the minimum possible money needed to ensure the security of all the junctions. And the second one is the number of ways you can ensure the security modulo 1000000007 (109 + 7).

Examples

input

31 2 331 22 33 2

output

3 1

input

52 8 0 6 061 41 32 43 44 55 1

output

8 2

input

101 3 2 2 1 3 1 4 10 10121 22 33 13 44 55 65 76 47 38 99 1010 9

output

15 6

input

27 9121 22 1

output

7 1

强连通分量模板题。

#include<iostream>#include<stdio.h>#include<string.h>#include<vector>using namespace std;#define MAXX 0x3fffffff#define MAX_V 100005#define MOD  1000000007int V;vector<int>G[MAX_V];vector<int>rG[MAX_V];vector<int>vs;bool used[MAX_V];int cmp[MAX_V];int cost[MAX_V];__int64 minx;int num;__int64 total;void add_edge(int from,int to){G[from].push_back(to);rG[to].push_back(from); }void dfs(int v){used[v]=true;for(int i=0;i<G[v].size();i++){if(!used[G[v][i]])dfs(G[v][i]);}vs.push_back(v);}void rdfs(int v,int k){used[v]=true;cmp[v]=k;for(int i=0;i<rG[v].size();i++){if(!used[rG[v][i]])rdfs(rG[v][i],k);}if(cost[v]==minx){num++;}else if(cost[v]<minx){minx=cost[v];num=1;}}__int64 scc(){memset(used,0,sizeof(used));total=1;__int64 totcost=0;vs.clear();for(int v=0;v<V;v++){if(!used[v])dfs(v);}memset(used,0,sizeof(used));int k=0;for(int i=vs.size()-1;i>=0;i--){if(!used[vs[i]]){num=1;minx=MAXX;rdfs(vs[i],k++);totcost+=minx;total=(total*num)%MOD;}}return totcost;}int main(){int n,m;while(~scanf("%d",&n)){memset(cmp,0,sizeof(cmp));memset(cost,0,sizeof(cost));for(int i=0;i<n;i++){G[i].clear();rG[i].clear();}vs.clear();V=n;for(int i=0;i<n;i++){scanf("%d",&cost[i]);}scanf("%d",&m); int f,t;for(int i=0;i<m;i++){scanf("%d%d",&f,&t);add_edge(f-1,t-1);}__int64 ans=scc();printf("%I64d %I64d\n",ans,total);}}