Codeforces #406 (Div. 2) B. Bear and Friendship Condition（并查集

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本文介绍了一个社交网络中成员间朋友关系合理性的判断方法。通过使用并查集算法确定社交网络中成员之间的连通性，并根据节点与边的数量关系来判断网络是否合理。文章详细解释了算法的具体实现过程。

B. Bear and Friendship Condition

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).

There are n members, numbered 1 through n. m pairs of members are friends. Of course, a member can’t be a friend with themselves.

Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.

For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.

Can you help Limak and check if the network is reasonable? Print “YES” or “NO” accordingly, without the quotes.

Input

The first line of the input contain two integers n and m (3 ≤ n ≤ 150 000, $\text{[math]}$ ) — the number of members and the number of pairs of members that are friends.

The i-th of the next m lines contains two distinct integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i). Members a_i and b_i are friends with each other. No pair of members will appear more than once in the input.

Output

If the given network is reasonable, print “YES” in a single line (without the quotes). Otherwise, print “NO” in a single line (without the quotes).

Examples

Input

Output

YES

Input

Output

NO

Input

Output

YES

Input

3 2
1 2
2 3

Output

NO

Note

The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is “NO” in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.

$\text{[math]}$

并查集求联通分量，节点与边数的关系（还是蛮麻烦的2333

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define N 150000+100
int par[N];
LL num[N], arr[N];

void init(int n)
{
    for(int i = 1;i <= n; i++) {
        par[i] = i;
        num[i] = 1;
        arr[i] = 0;
    }
}

int find(int x)
{
    if(x == par[x]) return x;
    else return par[x]=find(par[x]);
}

int main()
{
    int n, m;
    scanf("%d%d",&n,&m);
    init(n);
    int x, y;
    for (int i = 0; i < m; i++){
        scanf("%d%d", &x, &y);
        if (find(x) == find(y)){ arr[par[x]]++; }
        else{
            num[par[y]] += num[par[x]];
            arr[par[y]] += arr[par[x]] + 1;
            par[par[x]] = par[y];
        }
    }
    LL sum ;
    bool flag = true;
    for(int i = 1;i <= n; i++) {
        if(find(i) == i) {
            sum = num[i] * (num[i] - 1) / 2;;
            if(arr[i] != sum) {
                flag = false;
                break;
            }
        }
    } 
    if(flag) puts("YES");
    else puts("NO");
return 0;
}