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>
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:20000000")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
constexpr int infValue =
    std::is_same<T, int>::value ? 2000100007 : 8600000000000000007ll;
template <typename T>
constexpr int maxValue =
    std::is_same<T, int>::value ? 1000000007 : 1000000000000000001ll;
const int INF = infValue<int>;
const long long MOD = 1000000007ll;
const double EPS = 1e-6;
const int MAX = maxValue<int>;
using pr = pair<int, int>;
int n, m;
int a, b;
vector<int> d[200010];
bool ok;
vector<int> z_function() {
  int n = ::n * 2;
  vector<int> z(n);
  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 && d[z[i]] == d[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
int main(void) {
  scanf("%d %d", &n, &m);
  while (m--) {
    scanf("%d %d", &a, &b), --a, --b;
    if (a > b) swap(a, b);
    int delta = b - a;
    d[a].push_back(delta), d[a + n].push_back(delta);
    delta = n - delta;
    d[b].push_back(delta), d[b + n].push_back(delta);
  }
  for (int i = 0; i < 2 * n; ++i) sort(d[i].begin(), d[i].end());
  auto z = z_function();
  for (int i = 0; i < n; ++i) ok |= (i + z[i] == n * 2);
  puts(ok ? "Yes" : "No");
  return 0;
}
