#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct kmp {
  int P;
  vector<T> word;
  vector<int> fail;
  int adv(int len, const T& nxt) const {
    while (len > 0 && word[len] != nxt) len = fail[len];
    return len + int(word[len] == nxt);
  }
  kmp(const vector<T>& word) : P(word.size()), word(word) {
    fail.resize(P + 1);
    for (int i = 2; i <= P; i++) fail[i] = adv(fail[i - 1], word[i - 1]);
  }
  vector<bool> find(const vector<T>& text) const {
    vector<bool> match(text.size());
    for (int i = 0, len = 0; i < text.size(); i++) {
      len = adv(len, text[i]);
      if (len == P) {
        match[i - len + 1] = true;
        len = fail[len];
      }
    }
    return match;
  }
};
const int MAXN = 101010;
int N, M;
vector<int> delta[MAXN];
void debug_vector(const vector<int>& v) {
  for (int i = 0; i < v.size(); ++i) {
    printf("%d ", v[i]);
  }
  printf("\n");
}
int main() {
  scanf("%d %d", &N, &M);
  for (int i = 0; i < M; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    delta[a].push_back((b - a + N) % N);
    delta[b].push_back((a - b + N) % N);
  }
  vector<int> str;
  for (int i = 0; i < N; ++i) {
    sort(delta[i].begin(), delta[i].end());
    str.push_back(-1);
    for (int j = 0; j < delta[i].size(); ++j) {
      str.push_back(delta[i][j]);
    }
  }
  vector<int> str2;
  for (int rep = 0; rep < 2; ++rep) {
    for (int i = 0; i < str.size(); ++i) {
      str2.push_back(str[i]);
    }
  }
  kmp<int> matcher(str);
  vector<bool> res = matcher.find(str2);
  for (int i = 1; i < str.size(); ++i) {
    if (res[i]) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
template <typename T>
struct kmp {
  int P;
  vector<T> word;
  vector<int> fail;
  int adv(int len, const T& nxt) const {
    while (len > 0 && word[len] != nxt) len = fail[len];
    return len + int(word[len] == nxt);
  }
  kmp(const vector<T>& word) : P(word.size()), word(word) {
    fail.resize(P + 1);
    for (int i = 2; i <= P; i++) fail[i] = adv(fail[i - 1], word[i - 1]);
  }
  vector<bool> find(const vector<T>& text) const {
    vector<bool> match(text.size());
    for (int i = 0, len = 0; i < text.size(); i++) {
      len = adv(len, text[i]);
      if (len == P) {
        match[i - len + 1] = true;
        len = fail[len];
      }
    }
    return match;
  }
};
const int MAXN = 101010;
int N, M;
vector<int> delta[MAXN];
void debug_vector(const vector<int>& v) {
  for (int i = 0; i < v.size(); ++i) {
    printf("%d ", v[i]);
  }
  printf("\n");
}
int main() {
  scanf("%d %d", &N, &M);
  for (int i = 0; i < M; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    delta[a].push_back((b - a + N) % N);
    delta[b].push_back((a - b + N) % N);
  }
  vector<int> str;
  for (int i = 0; i < N; ++i) {
    sort(delta[i].begin(), delta[i].end());
    str.push_back(-1);
    for (int j = 0; j < delta[i].size(); ++j) {
      str.push_back(delta[i][j]);
    }
  }
  vector<int> str2;
  for (int rep = 0; rep < 2; ++rep) {
    for (int i = 0; i < str.size(); ++i) {
      str2.push_back(str[i]);
    }
  }
  kmp<int> matcher(str);
  vector<bool> res = matcher.find(str2);
  for (int i = 1; i < str.size(); ++i) {
    if (res[i]) {
      printf("YES\n");
      return 0;
    }
  }
  printf("NO\n");
  return 0;
}

```