#include <bits/stdc++.h>
using namespace std;
int inf = 0x3f3f3f3f;
long long infl = 0x3f3f3f3f3f3f3f3fLL;
long double infd = 1.0 / 0.0;
const double pi = 2 * acos(0.0);
const int MOD = 1e9 + 7;
vector<pair<int, int>> pre;
vector<pair<int, int>> nrw;
const int mxx = 2 * 100 * 1007;
bool prime[mxx];
void seive() {
  for (int i = 0; i < mxx; i++) prime[i] = true;
  for (int i = 2; i < mxx; i++)
    if (prime[i])
      for (int j = 2 * i; j < mxx; j += i) prime[j] = false;
}
int main() {
  seive();
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    pre.push_back({a, b});
  }
  sort(pre.begin(), pre.end());
  for (int i = 2; i <= n; i++) {
    if (prime[i] && n % i == 0) {
      int x = n / i;
      nrw.clear();
      for (int j = 0; j < m; j++) {
        int a, b;
        a = (pre[j].first + x) > n ? (pre[j].first + x) % n
                                   : (pre[j].first + x);
        b = (pre[j].second + x) > n ? (pre[j].second + x) % n
                                    : (pre[j].second + x);
        if (a > b) swap(a, b);
        nrw.push_back({a, b});
      }
      sort(nrw.begin(), nrw.end());
      if (pre == nrw) return 0 * puts("YES");
    }
  }
  return 0 * puts("NO");
}

### Prompt

In CPP, your task is to solve 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 inf = 0x3f3f3f3f;
long long infl = 0x3f3f3f3f3f3f3f3fLL;
long double infd = 1.0 / 0.0;
const double pi = 2 * acos(0.0);
const int MOD = 1e9 + 7;
vector<pair<int, int>> pre;
vector<pair<int, int>> nrw;
const int mxx = 2 * 100 * 1007;
bool prime[mxx];
void seive() {
  for (int i = 0; i < mxx; i++) prime[i] = true;
  for (int i = 2; i < mxx; i++)
    if (prime[i])
      for (int j = 2 * i; j < mxx; j += i) prime[j] = false;
}
int main() {
  seive();
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    if (a > b) swap(a, b);
    pre.push_back({a, b});
  }
  sort(pre.begin(), pre.end());
  for (int i = 2; i <= n; i++) {
    if (prime[i] && n % i == 0) {
      int x = n / i;
      nrw.clear();
      for (int j = 0; j < m; j++) {
        int a, b;
        a = (pre[j].first + x) > n ? (pre[j].first + x) % n
                                   : (pre[j].first + x);
        b = (pre[j].second + x) > n ? (pre[j].second + x) % n
                                    : (pre[j].second + x);
        if (a > b) swap(a, b);
        nrw.push_back({a, b});
      }
      sort(nrw.begin(), nrw.end());
      if (pre == nrw) return 0 * puts("YES");
    }
  }
  return 0 * puts("NO");
}

```