做法很简单，2~n枚举点，每次建并查集搞搞，除了枚举的点以外都连，然后查一下1和n在不在一个集合，不在就说明枚举的点是一定经过的点

#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;
inline int read()
{
    register int num = 0;
    register char ch;
    register bool flag = false;
    while ((ch = getchar()) == ' ' || ch == '\n' || ch == '\r');
    if (ch == '-')flag = true; else num = ch ^ 48;
    while ((ch = getchar()) != ' '&&ch != '\n'&&ch != '\r'&&~ch)
        num = (num << 1) + (num << 3) + (ch ^ 48);
    if (flag)return -num; return num;
}
int fa[N],ans[N];
struct Node {
    int x, y;
}edge[N];
inline int findf(register int x)
{
    if (fa[x] == x)return x;
    return fa[x] = findf(fa[x]);
}

int main()
{
    register int n = read(), m = read();
    for (register int i = 1; i <= m; ++i)edge[i].x = read(), edge[i].y = read();
    for (register int i = 2; i != n; ++i)
    {
        for (register int j = 1; j <= n; ++j)fa[j] = j;
        for (register int j = 1; j <= m; ++j)
        {
            register int fx = findf(edge[j].x), fy = findf(edge[j].y);
            if (edge[j].x != i && edge[j].y != i && fx != fy)fa[fx] = fy;
        }
        if (findf(1) != findf(n))ans[++ans[0]] = i;
    }
    printf("%d\n", ans[0]);
    for (register int i = 1; i <= ans[0]; ++i)printf("%d ", ans[i]);
    return 0;
}