#include <bits/stdc++.h>
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(pair<long long, long long> a) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(a.first + FIXED_RANDOM) +
           splitmix64(a.second + FIXED_RANDOM);
  }
};
vector<long long> factorize(long long num) {
  vector<long long> res;
  for (long long i = 1; i <= sqrt(num); i++) {
    if ((num % i) == 0) {
      if (i * i == num) {
        res.push_back(i);
      } else {
        res.push_back(i);
        res.push_back(num / i);
      }
    }
  }
  sort(res.begin(), res.end());
  return res;
}
int main() {
  long long(n), (m);
  scanf("%lld%lld", &(n), &(m));
  ;
  vector<pair<long long, long long> > A, B;
  unordered_set<pair<long long, long long>, custom_hash> second;
  for (long long i = 0; i < (m); i++) {
    long long(a), (b);
    scanf("%lld%lld", &(a), &(b));
    ;
    --a;
    --b;
    A.push_back({a, b});
    second.insert({a, b});
  }
  vector<long long> factors = factorize(n);
  for (auto i : factors) {
    if (i == n) continue;
    bool flag = 0;
    for (long long j = 0; j < (m); j++) {
      long long a = (A[j].first + i) % n;
      long long b = (A[j].second + i) % n;
      if (second.find({a, b}) == second.end() &&
          second.find({b, a}) == second.end()) {
        flag = 1;
        break;
      }
    }
    if (!flag) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

### 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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(pair<long long, long long> a) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(a.first + FIXED_RANDOM) +
           splitmix64(a.second + FIXED_RANDOM);
  }
};
vector<long long> factorize(long long num) {
  vector<long long> res;
  for (long long i = 1; i <= sqrt(num); i++) {
    if ((num % i) == 0) {
      if (i * i == num) {
        res.push_back(i);
      } else {
        res.push_back(i);
        res.push_back(num / i);
      }
    }
  }
  sort(res.begin(), res.end());
  return res;
}
int main() {
  long long(n), (m);
  scanf("%lld%lld", &(n), &(m));
  ;
  vector<pair<long long, long long> > A, B;
  unordered_set<pair<long long, long long>, custom_hash> second;
  for (long long i = 0; i < (m); i++) {
    long long(a), (b);
    scanf("%lld%lld", &(a), &(b));
    ;
    --a;
    --b;
    A.push_back({a, b});
    second.insert({a, b});
  }
  vector<long long> factors = factorize(n);
  for (auto i : factors) {
    if (i == n) continue;
    bool flag = 0;
    for (long long j = 0; j < (m); j++) {
      long long a = (A[j].first + i) % n;
      long long b = (A[j].second + i) % n;
      if (second.find({a, b}) == second.end() &&
          second.find({b, a}) == second.end()) {
        flag = 1;
        break;
      }
    }
    if (!flag) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}

```