#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long MOD = 1e9 + 7;
inline long long add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline long long mul(long long x, long long y) { return (x * 1ll * y) % mod; }
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 2e5 + 5;
vector<pair<long long, long long>> v;
long long l;
vector<bool> prime(N + 5, true);
vector<long long> pr;
void sieve() {
  for (long long i = 2; i <= N; i++) {
    if (prime[i]) {
      for (long long j = 2 * i; j <= N; j += i) {
        prime[j] = false;
      }
    }
  }
  for (long long i = 2; i <= N; i++) {
    if (prime[i]) {
      pr.push_back(i);
    }
  }
}
bool check(long long n) {
  vector<pair<long long, long long>> k;
  long long x, y;
  for (long long i = 0; i < v.size(); i++) {
    x = (v[i].first + n) % l, y = (v[i].second + n) % l;
    if (x > y) swap(x, y);
    k.push_back({x, y});
  }
  sort(k.begin(), k.end());
  if (k == v) return true;
  return false;
}
signed main() {
  {
    long long n, m;
    cin >> n >> m;
    long long x, y;
    l = n;
    for (long long i = 0; i < m; i++) {
      cin >> x >> y;
      x %= l, y %= l;
      if (x > y) swap(x, y);
      v.push_back({x, y});
    }
    sort(v.begin(), v.end());
    sieve();
    for (auto it : pr) {
      if (n % it == 0) {
        if (check(n / it)) return cout << "Yes", 0;
      }
    }
    cout << "NO";
  }
}

### Prompt

Generate a cpp solution 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;
const long long mod = 1e9 + 7;
const long long MOD = 1e9 + 7;
inline long long add(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline long long mul(long long x, long long y) { return (x * 1ll * y) % mod; }
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 2e5 + 5;
vector<pair<long long, long long>> v;
long long l;
vector<bool> prime(N + 5, true);
vector<long long> pr;
void sieve() {
  for (long long i = 2; i <= N; i++) {
    if (prime[i]) {
      for (long long j = 2 * i; j <= N; j += i) {
        prime[j] = false;
      }
    }
  }
  for (long long i = 2; i <= N; i++) {
    if (prime[i]) {
      pr.push_back(i);
    }
  }
}
bool check(long long n) {
  vector<pair<long long, long long>> k;
  long long x, y;
  for (long long i = 0; i < v.size(); i++) {
    x = (v[i].first + n) % l, y = (v[i].second + n) % l;
    if (x > y) swap(x, y);
    k.push_back({x, y});
  }
  sort(k.begin(), k.end());
  if (k == v) return true;
  return false;
}
signed main() {
  {
    long long n, m;
    cin >> n >> m;
    long long x, y;
    l = n;
    for (long long i = 0; i < m; i++) {
      cin >> x >> y;
      x %= l, y %= l;
      if (x > y) swap(x, y);
      v.push_back({x, y});
    }
    sort(v.begin(), v.end());
    sieve();
    for (auto it : pr) {
      if (n % it == 0) {
        if (check(n / it)) return cout << "Yes", 0;
      }
    }
    cout << "NO";
  }
}

```