#include <bits/stdc++.h>
using namespace std;
int n, m;
struct seg {
  int l, r, d;
  inline bool operator<(const seg& x) const {
    return d < x.d || d == x.d && l < x.l;
  }
  inline bool operator!=(const seg& x) const {
    return l != x.l || r != x.r || d != x.d;
  }
} a[200005], b[200005];
inline bool check(int d) {
  for (int i = 1; i <= m; i++) {
    b[i].l = (a[i].l + d - 1) % n + 1;
    b[i].r = (a[i].r + d - 1) % n + 1;
    b[i].d = a[i].d;
    if (b[i].d * 2 == n && b[i].l > b[i].r) swap(b[i].l, b[i].r);
  }
  sort(b + 1, b + m + 1);
  for (int i = 1; i <= m; i++)
    if (a[i] != b[i]) return false;
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    a[i].d = (a[i].r - a[i].l + n) % n;
    if (a[i].d * 2 > n) {
      swap(a[i].l, a[i].r);
      a[i].d = n - a[i].d;
    } else if (a[i].d * 2 == n) {
      if (a[i].l > a[i].r) swap(a[i].l, a[i].r);
    }
  }
  sort(a + 1, a + m + 1);
  for (int d = 1; d * d <= n; d++) {
    if (n % d) continue;
    if (check(d)) {
      puts("Yes");
      return 0;
    }
    if (d > 1 && check(n / d)) {
      puts("Yes");
      return 0;
    }
  }
  puts("No");
  return 0;
}

### Prompt

Your challenge is to write 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;
int n, m;
struct seg {
  int l, r, d;
  inline bool operator<(const seg& x) const {
    return d < x.d || d == x.d && l < x.l;
  }
  inline bool operator!=(const seg& x) const {
    return l != x.l || r != x.r || d != x.d;
  }
} a[200005], b[200005];
inline bool check(int d) {
  for (int i = 1; i <= m; i++) {
    b[i].l = (a[i].l + d - 1) % n + 1;
    b[i].r = (a[i].r + d - 1) % n + 1;
    b[i].d = a[i].d;
    if (b[i].d * 2 == n && b[i].l > b[i].r) swap(b[i].l, b[i].r);
  }
  sort(b + 1, b + m + 1);
  for (int i = 1; i <= m; i++)
    if (a[i] != b[i]) return false;
  return true;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    a[i].d = (a[i].r - a[i].l + n) % n;
    if (a[i].d * 2 > n) {
      swap(a[i].l, a[i].r);
      a[i].d = n - a[i].d;
    } else if (a[i].d * 2 == n) {
      if (a[i].l > a[i].r) swap(a[i].l, a[i].r);
    }
  }
  sort(a + 1, a + m + 1);
  for (int d = 1; d * d <= n; d++) {
    if (n % d) continue;
    if (check(d)) {
      puts("Yes");
      return 0;
    }
    if (d > 1 && check(n / d)) {
      puts("Yes");
      return 0;
    }
  }
  puts("No");
  return 0;
}

```