#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <bit>
#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define bw(x) bit_width(unsigned(x))
#define no() { cout << "no\n"; return 0; }

using namespace std;

int main() {
    cin.tie(0)->sync_with_stdio(0);

    int n, m;
    cin >> n >> m;

    vector<vector<int>> adj(n);
    rep(i, 0, m) {
        int a, b; cin >> a >> b;
        --a, --b;
        adj[a].emplace_back(b);
        adj[b].emplace_back(a);
    }

    int d = bw(n) - 1;
    if (!(n == (1 << d) && ranges::all_of(adj, [d](auto& v) { return v.size() == d; }))) no()

    vector visited(n, false), inQ(n, false);
    vector value(n, 0);
    visited[0] = true;
    queue<int> q;
    rep(i, 0, d) {
        value[adj[0][i]] = 1 << i;
        inQ[adj[0][i]] = true;
        q.push(adj[0][i]);
    }
    for (; !q.empty(); q.pop()) {
        int u = q.front();
        visited[u] = true;
        for (auto v : adj[u]) {
            if (visited[v]) continue;
            // In a hypercube, we know that the distance
            // to 0 is the same as the number of bits set.
            // Also, the bitwise XOR of adjacent nodes is 1.
            // So u is always a bitwise subset of v.
            value[v] |= value[u];
            if (!inQ[v]) q.push(v), inQ[v] = true;
        }
    }

    rep(u, 0, n)
        for (auto v : adj[u])
            if (popcount(unsigned(value[u] ^ value[v])) != 1) no()

    cout << "yes\n";
}