#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int inf = 1 << 30;
const int maxn = 200000 + 5;
int n, m;
pair<int, int> a[maxn], b[maxn];
bool check(int x) {
  if (x == n) return 0;
  for (int i = 1; i <= m; i++) {
    b[i].first = (a[i].first + x - 1) % n + 1;
    b[i].second = (a[i].second + x - 1) % n + 1;
    if (b[i].first > b[i].second) swap(b[i].first, b[i].second);
  }
  sort(b + 1, b + 1 + m);
  for (int i = 1; i <= m; i++)
    if (a[i] != b[i]) return 0;
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].first, &a[i].second);
    if (a[i].first > a[i].second) swap(a[i].first, a[i].second);
  }
  sort(a + 1, a + 1 + m);
  for (int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    if (check(i)) return puts("Yes"), 0;
    if (check(n / i)) return puts("Yes"), 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;
const int mod = 998244353;
const int inf = 1 << 30;
const int maxn = 200000 + 5;
int n, m;
pair<int, int> a[maxn], b[maxn];
bool check(int x) {
  if (x == n) return 0;
  for (int i = 1; i <= m; i++) {
    b[i].first = (a[i].first + x - 1) % n + 1;
    b[i].second = (a[i].second + x - 1) % n + 1;
    if (b[i].first > b[i].second) swap(b[i].first, b[i].second);
  }
  sort(b + 1, b + 1 + m);
  for (int i = 1; i <= m; i++)
    if (a[i] != b[i]) return 0;
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].first, &a[i].second);
    if (a[i].first > a[i].second) swap(a[i].first, a[i].second);
  }
  sort(a + 1, a + 1 + m);
  for (int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    if (check(i)) return puts("Yes"), 0;
    if (check(n / i)) return puts("Yes"), 0;
  }
  puts("No");
  return 0;
}

```