#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result *= a;
    }
    a *= a;
    b /= 2;
  }
  return result;
}
long long gcd(long long x, long long y) {
  long long r;
  while (y != 0 && (r = x % y) != 0) {
    x = y;
    y = r;
  }
  return y == 0 ? x : y;
}
long long countSetBits(long long x) {
  long long Count = 0;
  while (x > 0) {
    if (x & 1) Count++;
    x = x >> 1;
  }
  return Count;
}
bool isPerfectSquare(long long n) {
  long long sr = sqrt(n);
  if (sr * sr == n)
    return true;
  else
    return false;
}
long long mod(long long x, long long M) { return ((x % M + M) % M); }
long long add(long long a, long long b, long long M) {
  return mod(mod(a, M) + mod(b, M), M);
}
long long mul(long long a, long long b, long long M) {
  return mod(mod(a, M) * mod(b, M), M);
}
long long powerM(long long a, long long b, long long M) {
  long long res = 1ll;
  while (b) {
    if (b % 2ll == 1ll) {
      res = mul(a, res, M);
    }
    a = mul(a, a, M);
    b /= 2ll;
  }
  return res;
}
long long mod_inv(long long a, long long m) {
  long long g = m, r = a, x = 0, y = 1;
  while (r != 0) {
    long long q = g / r;
    g %= r;
    swap(g, r);
    x -= q * y;
    swap(x, y);
  }
  return mod(x, m);
}
long long nCr(long long n, long long k) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long modInverse(long long n, long long M) { return powerM(n, M - 2, M); }
vector<long long> factu;
void factcalc(long long n, long long M) {
  factu.resize(n + 1);
  factu[0] = 1;
  for (long long i = 1; i <= n; i++) {
    factu[i] = mul(factu[i - 1], i, M);
  }
}
long long nCrM(long long n, long long r, long long M) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return mul(mul(factu[n], modInverse(factu[r], M), M),
             modInverse(factu[n - r], M), M);
}
long long rand_int(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long> > v(n);
  for (long long i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    v[x - 1].push_back(mod(y - x, n));
    v[y - 1].push_back(mod(x - y, n));
  }
  for (long long i = 0; i < n; i++) {
    v[i].push_back(0);
    sort(v[i].begin(), v[i].end());
  }
  vector<long long> v2;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < v[i].size(); j++) {
      v2.push_back(v[i][j]);
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < v[i].size(); j++) {
      v2.push_back(v[i][j]);
    }
  }
  n = v2.size() / 2;
  auto pfunc = [&](vector<long long> a) -> vector<long long> {
    long long nn = a.size();
    vector<long long> ret(nn);
    ret[0] = 0;
    for (long long i = 1; i < nn; i++) {
      long long op = ret[i - 1];
      while (op > 0 && a[i] != a[op]) {
        op = ret[op - 1];
      }
      if (a[i] == a[op]) {
        op++;
      }
      ret[i] = op;
    }
    return ret;
  };
  vector<long long> pref_n = pfunc(v2);
  for (long long i = n; i < 2 * n - 1; i++) {
    if (pref_n[i] == n) {
      cout << "Yes\n";
      return;
    }
  }
  cout << "No\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(20);
  solve();
  return 0;
}

### Prompt

Your challenge is to write 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 double PI = acos(-1.0);
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result *= a;
    }
    a *= a;
    b /= 2;
  }
  return result;
}
long long gcd(long long x, long long y) {
  long long r;
  while (y != 0 && (r = x % y) != 0) {
    x = y;
    y = r;
  }
  return y == 0 ? x : y;
}
long long countSetBits(long long x) {
  long long Count = 0;
  while (x > 0) {
    if (x & 1) Count++;
    x = x >> 1;
  }
  return Count;
}
bool isPerfectSquare(long long n) {
  long long sr = sqrt(n);
  if (sr * sr == n)
    return true;
  else
    return false;
}
long long mod(long long x, long long M) { return ((x % M + M) % M); }
long long add(long long a, long long b, long long M) {
  return mod(mod(a, M) + mod(b, M), M);
}
long long mul(long long a, long long b, long long M) {
  return mod(mod(a, M) * mod(b, M), M);
}
long long powerM(long long a, long long b, long long M) {
  long long res = 1ll;
  while (b) {
    if (b % 2ll == 1ll) {
      res = mul(a, res, M);
    }
    a = mul(a, a, M);
    b /= 2ll;
  }
  return res;
}
long long mod_inv(long long a, long long m) {
  long long g = m, r = a, x = 0, y = 1;
  while (r != 0) {
    long long q = g / r;
    g %= r;
    swap(g, r);
    x -= q * y;
    swap(x, y);
  }
  return mod(x, m);
}
long long nCr(long long n, long long k) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long modInverse(long long n, long long M) { return powerM(n, M - 2, M); }
vector<long long> factu;
void factcalc(long long n, long long M) {
  factu.resize(n + 1);
  factu[0] = 1;
  for (long long i = 1; i <= n; i++) {
    factu[i] = mul(factu[i - 1], i, M);
  }
}
long long nCrM(long long n, long long r, long long M) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return mul(mul(factu[n], modInverse(factu[r], M), M),
             modInverse(factu[n - r], M), M);
}
long long rand_int(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long> > v(n);
  for (long long i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    v[x - 1].push_back(mod(y - x, n));
    v[y - 1].push_back(mod(x - y, n));
  }
  for (long long i = 0; i < n; i++) {
    v[i].push_back(0);
    sort(v[i].begin(), v[i].end());
  }
  vector<long long> v2;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < v[i].size(); j++) {
      v2.push_back(v[i][j]);
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < v[i].size(); j++) {
      v2.push_back(v[i][j]);
    }
  }
  n = v2.size() / 2;
  auto pfunc = [&](vector<long long> a) -> vector<long long> {
    long long nn = a.size();
    vector<long long> ret(nn);
    ret[0] = 0;
    for (long long i = 1; i < nn; i++) {
      long long op = ret[i - 1];
      while (op > 0 && a[i] != a[op]) {
        op = ret[op - 1];
      }
      if (a[i] == a[op]) {
        op++;
      }
      ret[i] = op;
    }
    return ret;
  };
  vector<long long> pref_n = pfunc(v2);
  for (long long i = n; i < 2 * n - 1; i++) {
    if (pref_n[i] == n) {
      cout << "Yes\n";
      return;
    }
  }
  cout << "No\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(20);
  solve();
  return 0;
}

```