#include <bits/stdc++.h>
using namespace std;
const long long llINF = 9223372036854775807;
const int INF = 2147483647;
const int maxn = 2e5 + 7;
const int maxm = 5e4 + 7;
const long long mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-12;
long long n, m;
bool used[maxn];
pair<int, int> Segment[maxn];
vector<long long> d;
set<pair<int, int> > s;
void divide(long long x) {
  for (long long i = 1; i * i <= n; i++) {
    if (x % i == 0) d.push_back(i);
    if (i * i != x) d.push_back(x / i);
  }
}
int get_nxt(int x, int k) { return x + k > n ? (x + k) % n : x + k; }
int main(int argc, char const *argv[]) {
  scanf("%lld%lld", &n, &m);
  divide(n);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &Segment[i].first, &Segment[i].second);
    if (Segment[i].first > Segment[i].second)
      swap(Segment[i].first, Segment[i].second);
    s.insert(Segment[i]);
  }
  for (int kk = 0; kk < d.size(); kk++) {
    long long k = d[kk];
    if (k == n) continue;
    bool flag = true;
    for (int i = 1; i <= m; i++) {
      pair<int, int> nxt_par = make_pair(get_nxt(Segment[i].first, k),
                                         get_nxt(Segment[i].second, k));
      if (nxt_par.first > nxt_par.second) swap(nxt_par.first, nxt_par.second);
      if (!s.count(nxt_par)) {
        flag = false;
        break;
      }
    }
    if (flag) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

### Prompt

Please create a solution in CPP 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 llINF = 9223372036854775807;
const int INF = 2147483647;
const int maxn = 2e5 + 7;
const int maxm = 5e4 + 7;
const long long mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-12;
long long n, m;
bool used[maxn];
pair<int, int> Segment[maxn];
vector<long long> d;
set<pair<int, int> > s;
void divide(long long x) {
  for (long long i = 1; i * i <= n; i++) {
    if (x % i == 0) d.push_back(i);
    if (i * i != x) d.push_back(x / i);
  }
}
int get_nxt(int x, int k) { return x + k > n ? (x + k) % n : x + k; }
int main(int argc, char const *argv[]) {
  scanf("%lld%lld", &n, &m);
  divide(n);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &Segment[i].first, &Segment[i].second);
    if (Segment[i].first > Segment[i].second)
      swap(Segment[i].first, Segment[i].second);
    s.insert(Segment[i]);
  }
  for (int kk = 0; kk < d.size(); kk++) {
    long long k = d[kk];
    if (k == n) continue;
    bool flag = true;
    for (int i = 1; i <= m; i++) {
      pair<int, int> nxt_par = make_pair(get_nxt(Segment[i].first, k),
                                         get_nxt(Segment[i].second, k));
      if (nxt_par.first > nxt_par.second) swap(nxt_par.first, nxt_par.second);
      if (!s.count(nxt_par)) {
        flag = false;
        break;
      }
    }
    if (flag) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

```