#include <bits/stdc++.h>
using namespace std;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
inline lli mpow(lli a, lli b, lli m = 1000000007) {
  lli ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    a = (a * a) % m;
    b /= 2;
  }
  return ans;
}
int n, m, arr[100010];
lli h1[100010], h2[100010];
lli ohash1[100010], thash1[100010], ohash2[100010], thash2[100010];
vvi adj(100010);
void calc(lli hash1[], lli hash2[], vvi& adj) {
  for (int i = 1; i < n + 1; i++) {
    hash1[i] = 0;
    hash2[i] = 0;
    for (auto& x : adj[i]) {
      hash1[i] += h1[x];
      hash2[i] += h2[x];
      if (hash1[i] >= 1000000007) hash1[i] -= 1000000007;
      if (hash2[i] >= 1000000007) hash2[i] -= 1000000007;
    }
  }
}
bool check(int off) {
  for (int i = 1; i < n + 1; i++) arr[i] = (i + off - 1) % n + 1;
  vvi tadj(n + 2);
  for (int i = 1; i < n + 1; i++) tadj[arr[i]] = adj[i];
  for (int i = 1; i < n + 1; i++) {
    for (auto& x : tadj[i]) x = arr[x];
  }
  calc(thash1, thash2, tadj);
  for (int i = 1; i < n + 1; i++) {
    if (thash1[i] != ohash1[i]) return 0;
    if (thash2[i] != ohash2[i]) return 0;
  }
  return 1;
}
int tejas_919(int kkkk) {
  for (int i = 0; i < 100010; i++) h1[i] = (mpow(167, i) * i) % 1000000007;
  for (int i = 0; i < 100010; i++) h2[i] = mpow(367, i);
  lli k, q, u, v, temp = 0, ans = 0;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  calc(ohash1, ohash2, adj);
  for (int i = 1; i < n; i++) {
    if (n % i == 0 && check(i)) return cout << "Yes" << '\n', 0;
  }
  cout << "No" << '\n';
  return 0;
}
signed main() {
  if (!0) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    tejas_919(i + 1);
  }
  {};
}

### Prompt

Please create a solution in cpp to the following problem:

Inaka has a disc, the circumference of which is n units. The circumference is equally divided by n points numbered clockwise from 1 to n, such that points i and i + 1 (1 ≤ i < n) are adjacent, and so are points n and 1.

There are m straight segments on the disc, the endpoints of which are all among the aforementioned n points.

Inaka wants to know if her image is rotationally symmetrical, i.e. if there is an integer k (1 ≤ k < n), such that if all segments are rotated clockwise around the center of the circle by k units, the new image will be the same as the original one.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 200 000) — the number of points and the number of segments, respectively.

The i-th of the following m lines contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) that describe a segment connecting points a_i and b_i.

It is guaranteed that no segments coincide.

Output

Output one line — "Yes" if the image is rotationally symmetrical, and "No" otherwise (both excluding quotation marks).

You can output each letter in any case (upper or lower).

Examples

Input


12 6
1 3
3 7
5 7
7 11
9 11
11 3


Output


Yes


Input


9 6
4 5
5 6
7 8
8 9
1 2
2 3


Output


Yes


Input


10 3
1 2
3 2
7 2


Output


No


Input


10 2
1 6
2 7


Output


Yes

Note

The first two examples are illustrated below. Both images become the same as their respective original ones after a clockwise rotation of 120 degrees around the center.

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
inline lli mpow(lli a, lli b, lli m = 1000000007) {
  lli ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    a = (a * a) % m;
    b /= 2;
  }
  return ans;
}
int n, m, arr[100010];
lli h1[100010], h2[100010];
lli ohash1[100010], thash1[100010], ohash2[100010], thash2[100010];
vvi adj(100010);
void calc(lli hash1[], lli hash2[], vvi& adj) {
  for (int i = 1; i < n + 1; i++) {
    hash1[i] = 0;
    hash2[i] = 0;
    for (auto& x : adj[i]) {
      hash1[i] += h1[x];
      hash2[i] += h2[x];
      if (hash1[i] >= 1000000007) hash1[i] -= 1000000007;
      if (hash2[i] >= 1000000007) hash2[i] -= 1000000007;
    }
  }
}
bool check(int off) {
  for (int i = 1; i < n + 1; i++) arr[i] = (i + off - 1) % n + 1;
  vvi tadj(n + 2);
  for (int i = 1; i < n + 1; i++) tadj[arr[i]] = adj[i];
  for (int i = 1; i < n + 1; i++) {
    for (auto& x : tadj[i]) x = arr[x];
  }
  calc(thash1, thash2, tadj);
  for (int i = 1; i < n + 1; i++) {
    if (thash1[i] != ohash1[i]) return 0;
    if (thash2[i] != ohash2[i]) return 0;
  }
  return 1;
}
int tejas_919(int kkkk) {
  for (int i = 0; i < 100010; i++) h1[i] = (mpow(167, i) * i) % 1000000007;
  for (int i = 0; i < 100010; i++) h2[i] = mpow(367, i);
  lli k, q, u, v, temp = 0, ans = 0;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  calc(ohash1, ohash2, adj);
  for (int i = 1; i < n; i++) {
    if (n % i == 0 && check(i)) return cout << "Yes" << '\n', 0;
  }
  cout << "No" << '\n';
  return 0;
}
signed main() {
  if (!0) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    tejas_919(i + 1);
  }
  {};
}

```