#include <bits/stdc++.h>
using namespace std;
int fpb(int x, int y) {
  if (y) return fpb(y, x % y);
  return x;
}
int mod;
bool comp(pair<int, int> a, pair<int, int> b) {
  if (a.first < b.first) return true;
  if (a.first > b.first) return false;
  if (a.second % mod < b.second % mod) return true;
  if (a.second % mod > b.second % mod) return false;
  return a.second < b.second;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int shred = n;
  vector<pair<int, int> > ab;
  vector<int> occ(n, 0);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    if (a > b) swap(a, b);
    if ((a << 1) + n == (b << 1)) {
      shred = 2;
      continue;
    }
    if (b - a < a + n - b) {
      ab.push_back(make_pair(b - a, a));
    } else {
      ab.push_back(make_pair(a + n - b, b));
    }
    occ[ab.back().first]++;
  }
  m = ab.size();
  for (int i = 0; i < n; i++) {
    if (occ[i]) {
      shred = fpb(shred, occ[i]);
    }
  }
  if (shred == 1) {
    printf("NO\n");
    return 0;
  }
  bool found = false;
  for (int i = 2; i <= shred; i++) {
    if (shred % i == 0) {
      bool work = true;
      mod = n / i;
      sort(ab.begin(), ab.end(), comp);
      for (int j = 0; j < m; j++) {
        if (j % i != i - 1) {
          if (ab[j].first != ab[j + 1].first ||
              ab[j].second + mod != ab[j + 1].second) {
            work = false;
            break;
          }
        }
      }
      if (work) {
        found = true;
        break;
      }
      while (shred % i == 0) shred /= i;
    }
  }
  if (found) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
}

### Prompt

Create a solution in Cpp for 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;
int fpb(int x, int y) {
  if (y) return fpb(y, x % y);
  return x;
}
int mod;
bool comp(pair<int, int> a, pair<int, int> b) {
  if (a.first < b.first) return true;
  if (a.first > b.first) return false;
  if (a.second % mod < b.second % mod) return true;
  if (a.second % mod > b.second % mod) return false;
  return a.second < b.second;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int shred = n;
  vector<pair<int, int> > ab;
  vector<int> occ(n, 0);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    if (a > b) swap(a, b);
    if ((a << 1) + n == (b << 1)) {
      shred = 2;
      continue;
    }
    if (b - a < a + n - b) {
      ab.push_back(make_pair(b - a, a));
    } else {
      ab.push_back(make_pair(a + n - b, b));
    }
    occ[ab.back().first]++;
  }
  m = ab.size();
  for (int i = 0; i < n; i++) {
    if (occ[i]) {
      shred = fpb(shred, occ[i]);
    }
  }
  if (shred == 1) {
    printf("NO\n");
    return 0;
  }
  bool found = false;
  for (int i = 2; i <= shred; i++) {
    if (shred % i == 0) {
      bool work = true;
      mod = n / i;
      sort(ab.begin(), ab.end(), comp);
      for (int j = 0; j < m; j++) {
        if (j % i != i - 1) {
          if (ab[j].first != ab[j + 1].first ||
              ab[j].second + mod != ab[j + 1].second) {
            work = false;
            break;
          }
        }
      }
      if (work) {
        found = true;
        break;
      }
      while (shred % i == 0) shred /= i;
    }
  }
  if (found) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
}

```