#include <bits/stdc++.h>
using namespace std;
const long long INF64 = 1000000000000000000;
const long long BigPrime = 252097800623;
const long double LDINF = 1000000000000;
const long long Mmod = 998244353;
const long long INF = 1000000000;
const long long Mod = 1000000007;
const long double PI = acos(-1.);
const long double Eps = 10e-10;
const long long Prime = Mod;
const unsigned int N = 100005;
long long hashes[N], n, m;
vector<int> tmp[N];
int z[N];
void input() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    if (b < a) swap(a, b);
    tmp[a].push_back(b - a);
    tmp[b].push_back(a + n - b);
  }
  for (int i = 0; i < n; ++i) sort((tmp[i]).begin(), (tmp[i]).end());
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < (int)(tmp[i]).size(); ++j)
      hashes[i] = (hashes[i] * Prime + tmp[i][j]);
  }
}
void solve() {
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && hashes[z[i]] == hashes[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  for (int i = 0; i < n; ++i)
    if (z[i] == (n - i) && n % i == 0) {
      puts("Yes");
      return;
    }
  puts("No");
  return;
}
int main() {
  int testcases = 1;
  for (int _ = 0; _ < testcases; ++_) {
    input();
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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 long long INF64 = 1000000000000000000;
const long long BigPrime = 252097800623;
const long double LDINF = 1000000000000;
const long long Mmod = 998244353;
const long long INF = 1000000000;
const long long Mod = 1000000007;
const long double PI = acos(-1.);
const long double Eps = 10e-10;
const long long Prime = Mod;
const unsigned int N = 100005;
long long hashes[N], n, m;
vector<int> tmp[N];
int z[N];
void input() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    if (b < a) swap(a, b);
    tmp[a].push_back(b - a);
    tmp[b].push_back(a + n - b);
  }
  for (int i = 0; i < n; ++i) sort((tmp[i]).begin(), (tmp[i]).end());
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < (int)(tmp[i]).size(); ++j)
      hashes[i] = (hashes[i] * Prime + tmp[i][j]);
  }
}
void solve() {
  for (int i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && hashes[z[i]] == hashes[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  for (int i = 0; i < n; ++i)
    if (z[i] == (n - i) && n % i == 0) {
      puts("Yes");
      return;
    }
  puts("No");
  return;
}
int main() {
  int testcases = 1;
  for (int _ = 0; _ < testcases; ++_) {
    input();
    solve();
  }
  return 0;
}

```