Codeforces 427C Checkposts (强连通分量 tarjan模板)

Checkposts

time limit per test：2 seconds

memory limit per test：256 megabytes

Your city has n junctions. There arem 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 junctioni can protect junction j if either i = j or the police patrol car can go toj from i and then come back toi.

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 andin 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. Thei^th integer is the cost of building checkpost at thei^th junction (costs will be non-negative and will not exceed10⁹).

The next line will contain an integer m (0 ≤ m ≤ 3·10⁵). And each of the nextm lines contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u ≠ v). A pairu_i, v_i means, that there is a one-way road which goes fromu_i tov_i. 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 modulo1000000007 (10⁹ + 7).

Examples

Input

3
1 2 3
3
1 2
2 3
3 2

Output

3 1

Input

5
2 8 0 6 0
6
1 4
1 3
2 4
3 4
4 5
5 1

Output

8 2

Input

10
1 3 2 2 1 3 1 4 10 10
12
1 2
2 3
3 1
3 4
4 5
5 6
5 7
6 4
7 3
8 9
9 10
10 9

Output

15 6

Input

2
7 91
2
1 2
2 1

Output

7 1

题目链接：http://codeforces.com/problemset/problem/427/C

题目大意：给一个有向图，每个点一个权值，对每个强连通分量选一个点，求最小权值和和在权值和最小的情况下可选择的方案数

题目分析：上面的题目大意其实已经将原题意转换成图论问题，强连通分量直接tarjan搞，对每个取权值最小的然后判断有没有权值一样的，然后乘法计数原理

#include 
#include 
#include 
#include 
#define ll long long
using namespace std;
int const INF = 0x3fffffff;
int const MOD = 1e9 + 7;
int const MAX = 1e5 + 5;
int n, m, a[MAX];
int cur, dfn[MAX], low[MAX];
bool ins[MAX];
vector  vt[MAX];
stack  stk;
ll ans, tot;

void Tarjan(int u)
{
    dfn[u] = low[u] = ++cur;
    stk.push(u); 
    ins[u] = true;
    for(int i = 0; i < (int)vt[u].size(); i++) 
    {
        int v = vt[u][i];
        if(!dfn[v]) 
            Tarjan(v);
        if(ins[v]) 
            low[u] = min(low[u], low[v]);
    }
    if(dfn[u] == low[u]) 
    {
        int mi = INF, cnt = 0, v;
        do
        {
            v = stk.top(); 
            stk.pop();
            ins[v] = false;
            if(a[v] < mi) 
            { 
                mi = a[v]; 
                cnt = 0; 
            }
            if(a[v] == mi) 
                cnt ++;
        }while(u != v);
        ans += mi; 
        tot = ((tot % MOD) * (cnt % MOD)) % MOD;
    }
}

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    scanf("%d", &m);
    while(m--) 
    {
        int x, y; 
        scanf("%d %d", &x, &y);
        vt[x].push_back(y);
    }
    tot = 1;
    for(int i = 1; i <= n; i++)
        if(!dfn[i])
            Tarjan(i);
    printf("%I64d %I64d\n", ans, tot);
}