#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MAXN = 2e5 + 5;
struct HASH {
  size_t operator()(const pair<int, int>& x) const {
    return hash<long long>()(((long long)x.first) ^
                             (((long long)x.second) << 32));
  }
};
pair<int, int> edge[MAXN];
unordered_set<pair<int, int>, HASH> st;
void impossible() {
  cout << "No\n";
  exit(0);
}
void possible() {
  cout << "Yes\n";
  exit(0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    edge[i] = {min(u, v), max(u, v)};
    st.insert(edge[i]);
  }
  for (int k = 1; k < n; k++) {
    if (n % k) continue;
    bool flag = true;
    for (int i = 1; i <= m; i++) {
      int u = edge[i].first, v = edge[i].second;
      if (!st.count({(u + k) % n, (v + k) % n}) &&
          !st.count({(v + k) % n, (u + k) % n})) {
        flag = false;
        break;
      }
    }
    if (flag) possible();
  }
  impossible();
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MAXN = 2e5 + 5;
struct HASH {
  size_t operator()(const pair<int, int>& x) const {
    return hash<long long>()(((long long)x.first) ^
                             (((long long)x.second) << 32));
  }
};
pair<int, int> edge[MAXN];
unordered_set<pair<int, int>, HASH> st;
void impossible() {
  cout << "No\n";
  exit(0);
}
void possible() {
  cout << "Yes\n";
  exit(0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    edge[i] = {min(u, v), max(u, v)};
    st.insert(edge[i]);
  }
  for (int k = 1; k < n; k++) {
    if (n % k) continue;
    bool flag = true;
    for (int i = 1; i <= m; i++) {
      int u = edge[i].first, v = edge[i].second;
      if (!st.count({(u + k) % n, (v + k) % n}) &&
          !st.count({(v + k) % n, (u + k) % n})) {
        flag = false;
        break;
      }
    }
    if (flag) possible();
  }
  impossible();
}

```