题意：

解法：

显然答案一定是n的因子,那么枚举因子k,
判断k是否是答案即可.

check比较简单,
先将原线段存pair然后排序(预处理).
用vector存循环k步之后的线段,排序,比较vector是否相同即可.

ps:一开始我用map存pair,结果T了...
看到别人用set存pair+set.count()判断也过了.

复杂度应该是O(m*sq*log),sq是n的因子数.

code：

#include<bits/stdc++.h>
// #define SYNC_OFF
typedef std::vector<int> VE;
typedef std::pair<int,int> PI;
// #define int long long
#define ll long long
#define ull unsigned long long
//fast-coding
#define ST(x) x.begin()
#define ED(x) x.end()
#define RST(x) x.rbegin()
#define RED(x) x.end()
#define CL(x) x.clear();
#define all(a,n) a+1,a+1+n
#define ff(i,n) for(ll i=1;i<=n;i++)
#define rff(i,n) for(ll i=n;i>=1;i--)
#define fff(i,n) for(ll i=0;i<n;i++)
#define rfff(i,n) for(ll i=n-1;i>=0;i--)
#define SC(x) scanf("%s",x)
#define SL(x) strlen(x)
#define pss(a) push_back(a)
#define ps(a) push(a)
#define SZ(x) (int)x.size()
#define pee puts("");
#define eee putchar(' ');
#define re readdd()
#define pr(a) printtt(a)
int readdd(){int x=0,f=1;char c=getchar();//
while(!isdigit(c)&&c!='-')c=getchar();
if(c=='-')f=-1,c=getchar();
while(isdigit(c))x=x*10+c-'0',c=getchar();
return f*x;}
void printtt(int x){if(x<0)putchar('-'),x=-x;//
if(x>=10)printtt(x/10);putchar(x%10+'0');}
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}//
int ppow(int a,int b,int mod){a%=mod;//
int ans=1%mod;while(b){if(b&1)ans=(long long)ans*a%mod;
a=(long long)a*a%mod;b>>=1;}return ans;}
bool addd(int a,int b){return a>b;}
int lowbit(int x){return x&-x;}
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
bool isdigit(char c){return c>='0'&&c<='9';}
bool Isprime(int x){
    for(int i=2;i*i<=x;i++)if(x%i==0)return 0;
    return 1;
}
void ac(int x){if(x)puts("YES");else puts("NO");}
//
using namespace std;
// const int mod=998244353;
const int mod=1e9+7;
const int maxm=2e6+5;
vector<PI>g;
int n,m;
int get(int x){
    if(x>n)x-=n;
    return x;
}
bool check(int k){
    vector<PI>gg;
    fff(i,m){
        int l=get(g[i].first+k);
        int r=get(g[i].second+k);
        if(l>r)swap(l,r);
        gg.push_back({l,r});
    }
    sort(ST(gg),ED(gg));
    return g==gg;
}
void solve(){
    n=re,m=re;
    ff(i,m){
        int l=re,r=re;
        if(l>r)swap(l,r);
        g.push_back({l,r});
    }
    sort(ST(g),ED(g));
    ff(k,n-1){
        if(n%k)continue;
        if(check(k)){
            ac(1);return ;
        }
    }
    ac(0);
}
void Main(){
    // #define MULTI_CASE
    #ifdef MULTI_CASE
    int T;cin>>T;while(T--)
    #endif
    solve();
}
void Init(){
    #ifdef SYNC_OFF
    ios::sync_with_stdio(0);cin.tie(0);
    #endif
    #ifndef ONLINE_JUDGE
    freopen("../in.txt","r",stdin);
    freopen("../out.txt","w",stdout);
    #endif
}
signed main(){
    Init();
    Main();
    return 0;
}

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