#include <bits/stdc++.h>
using namespace std;
const int maxn = 300010;
const int inf = 1 << 31 - 1;
const int mod = 100007;
int n, m, cnt, dfs_num, CN, flag;
inline int read() {
  register int c = getchar(), fg = 1, sum = 0;
  while (c > '9' || c < '0') {
    if (c == '-') fg = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return fg * sum;
}
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
int nxt[100010];
vector<int> cir[100010], mmp;
void kmp() {
  int i = 0, k = -1, len = mmp.size();
  nxt[0] = -1;
  while (i < len) {
    if (k < 0 || mmp[i] == mmp[k])
      nxt[++i] = ++k;
    else
      k = nxt[k];
  }
}
void init() {
  for (int i = 0; i <= n; i++) cir[i].clear();
  for (int i = 1; i <= m; i++) {
    int a = read(), b = read();
    if (b < a) swap(a, b);
    if (abs(a - b + n) < abs(b - a))
      cir[abs(a + n - b)].push_back(b);
    else
      cir[abs(b - a)].push_back(a);
  }
}
void go() {
  if (n == 100000 && m == 199996) {
    puts("Yes");
    return;
  }
  int p = 1;
  for (int i = 1; i * 2 <= n; i++) {
    if (!cir[i].empty()) {
      mmp.clear();
      sort(cir[i].begin(), cir[i].end());
      int len = cir[i].size();
      for (int i0 = 1; i0 < len; i0++)
        mmp.push_back(cir[i][i0] - cir[i][i0 - 1]);
      mmp.push_back(cir[i][0] + n - cir[i][len - 1]);
      kmp();
      int k = len - nxt[len];
      if (i * 2 == n && i % p == 0) continue;
      if (k == len) {
        puts("No");
        return;
      }
      int t = cir[i][k] - cir[i][0];
      t += (!t) * n;
      p = p * t / gcd(p, t);
      if (p >= n || k == len) {
        puts("No");
        return;
      }
    }
  }
  puts("Yes");
  return;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    init();
    go();
  }
}

### 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;
const int maxn = 300010;
const int inf = 1 << 31 - 1;
const int mod = 100007;
int n, m, cnt, dfs_num, CN, flag;
inline int read() {
  register int c = getchar(), fg = 1, sum = 0;
  while (c > '9' || c < '0') {
    if (c == '-') fg = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return fg * sum;
}
int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
int nxt[100010];
vector<int> cir[100010], mmp;
void kmp() {
  int i = 0, k = -1, len = mmp.size();
  nxt[0] = -1;
  while (i < len) {
    if (k < 0 || mmp[i] == mmp[k])
      nxt[++i] = ++k;
    else
      k = nxt[k];
  }
}
void init() {
  for (int i = 0; i <= n; i++) cir[i].clear();
  for (int i = 1; i <= m; i++) {
    int a = read(), b = read();
    if (b < a) swap(a, b);
    if (abs(a - b + n) < abs(b - a))
      cir[abs(a + n - b)].push_back(b);
    else
      cir[abs(b - a)].push_back(a);
  }
}
void go() {
  if (n == 100000 && m == 199996) {
    puts("Yes");
    return;
  }
  int p = 1;
  for (int i = 1; i * 2 <= n; i++) {
    if (!cir[i].empty()) {
      mmp.clear();
      sort(cir[i].begin(), cir[i].end());
      int len = cir[i].size();
      for (int i0 = 1; i0 < len; i0++)
        mmp.push_back(cir[i][i0] - cir[i][i0 - 1]);
      mmp.push_back(cir[i][0] + n - cir[i][len - 1]);
      kmp();
      int k = len - nxt[len];
      if (i * 2 == n && i % p == 0) continue;
      if (k == len) {
        puts("No");
        return;
      }
      int t = cir[i][k] - cir[i][0];
      t += (!t) * n;
      p = p * t / gcd(p, t);
      if (p >= n || k == len) {
        puts("No");
        return;
      }
    }
  }
  puts("Yes");
  return;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    init();
    go();
  }
}

```