Problem:
Score : 400 points

Problem Statement

We have a permutation of the integers from 1 through N, p_1, p_2, .., p_N.
We also have M pairs of two integers between 1 and N (inclusive), represented as (x_1,y_1), (x_2,y_2), .., (x_M,y_M).
AtCoDeer the deer is going to perform the following operation on p as many times as desired so that the number of i (1 ≤ i ≤ N) such that p_i = i is maximized:

Choose j such that 1 ≤ j ≤ M, and swap p_{x_j} and p_{y_j}.

Find the maximum possible number of i such that p_i = i after operations.

Constraints

2 ≤ N ≤ 10^5

1 ≤ M ≤ 10^5

p is a permutation of integers from 1 through N.

1 ≤ x_j,y_j ≤ N

x_j ≠ y_j

If i ≠ j, \{x_i,y_i\} ≠ \{x_j,y_j\}.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
p_1 p_2 .. p_N
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the maximum possible number of i such that p_i = i after operations.

Sample Input 1

5 2
5 3 1 4 2
1 3
5 4

Sample Output 1

2

If we perform the operation by choosing j=1, p becomes 1 3 5 4 2, which is optimal, so the answer is 2.

Sample Input 2

3 2
3 2 1
1 2
2 3

Sample Output 2

3

If we perform the operation by, for example, choosing j=1, j=2, j=1 in this order, p becomes 1 2 3, which is obviously optimal.
Note that we may choose the same j any number of times.

Sample Input 3

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

Sample Output 3

8

Sample Input 4

5 1
1 2 3 4 5
1 5

Sample Output 4

5

We do not have to perform the operation.

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
map<int,int>mp;
vector<int>v[120000],res;
int n,m,p[120000],vis[120000];
inline void read(int &x)
{
    char c=getchar();x=0;
    while(c>'9'||c<'0')c=getchar();
    while(c<='9'&&c>='0') x=(x<<1)+(x<<3)+c-'0',c=getchar();
}
int x,y;
inline void dfs(int x,int f)
{
    vis[x]=1;
    res.push_back(x);
    for(int i=0;i<v[x].size();i++)
    {
        int y=v[x][i];
        if(y==f) continue;
        dfs(y,x);
    }
}
int ans=0;
int main()
{
    read(n);read(m);
    for(int i=1;i<=n;i++) read(p[i]);
    for(int i=1;i<=m;i++)
    {
        read(x),read(y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    for(int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
            res.clear();
            mp.clear();
            dfs(i,0);
            for(int j=0;j<res.size();j++)
                mp[res[j]]=1;
            
            for(int j=0;j<res.size();j++)
            {
                if(mp[p[res[j]]])  ans++;
            }
        }
    }
    printf("%d",ans);
    //system("pause");
}