Your task is so easy. I will give you an undirected graph, and you just need to tell me whether the graph is just a circle. A cycle is three or more nodesV₁, V₂,V₃, ... V_k, such that there are edges betweenV₁

Input

There are multiple cases (no more than 10).

The first line contains two integers n and m, which indicate the number of nodes and the number of edges (1 <n

Following are m lines, each contains two integers x and y (1 <= x, y <= n, x != y), which means there is an edge between nodex and node y.

There is a blank line between cases.

Output

If the graph is just a circle, output "YES", otherwise output "NO".

Sample Input

3 3
1 2
2 3
1 3

4 4
1 2
2 3
3 1
1 4

Sample Output

YES
NO

題目大意：給你n個結點m條邊組成無向圖，求該無向圖是不是一個環，是輸出“YES”， 否輸出“NO”；

分析：記錄每個節點的度【如果能是環，則每個節點的度均為2】，然後在判斷是否左右結點都在這個圖中；

可以在輸入邊時記錄該邊連線的兩個結點的度，判斷所有點是否在一個圖中，可以用並查集~~~

已Accept程式碼【c++】

#include <cstdio>
#include <cstring>
using namespace std;

int n, m;
int pre[11], map[11];

int Find(int x) {
	int r = x;
	while(r != pre[r])
		r = pre[r];
	int i = x, j;
	while(i != r) {
		j = pre[i];
		pre[i] = r;
		i = j;
	}
	return r;
}

void Join(int x, int y) {
	int fx = Find(x);
	int fy = Find(y);
	if(fx != fy)
		pre[fx] = fy;
}

bool Ok() {
	for(int i = 1; i <= n; i++)
		if(map[i] != 2)
			return false;
	int ok = Find(1);
	for(int i = 2; i <= n; i++)
		if(ok != Find(i))
			return false;
	return true;
}

int main() {
	int a, b;
	while(scanf("%d%d", &n, &m) != EOF) {
		for(int i = 1; i <= n; i++) {
			pre[i] = i;
			map[i] = 0;
		}
		for(int i = 0; i < m; i++) {
			scanf("%d %d", &a, &b);
			Join(a, b);
			map[a]++;
			map[b]++;
		}
		printf("%s\n", Ok() ? "YES" : "NO");
	}
	return 0;
}