#include <bits/stdc++.h>
using namespace std;
const int BUF = 1000000;
char buf[BUF], *p1, *p2;
inline char getChar() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, BUF, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  register int f = 0;
  register char c;
  while (!isdigit(c = getChar())) {
  }
  do f = f * 10 + (c ^ 48);
  while (isdigit(c = getChar()));
  return f;
}
const int N = 200003;
const long long yyb = 134874928973;
vector<int> g[N];
long long arr[N];
int fail[N];
int main() {
  int n = read(), m = read(), i, x, y;
  for (i = 1; i <= m; ++i) {
    x = read(), y = read();
    if (x > y) swap(x, y);
    g[x].push_back(y - x), g[y].push_back(x - y + n);
  }
  for (i = 1; i <= n; ++i) {
    sort(g[i].begin(), g[i].end());
    for (auto x : g[i]) arr[i] = (arr[i] * n + x) % yyb;
  }
  for (i = 1; i < n; ++i) arr[i + n] = arr[i];
  m = n << 1;
  bool ans = 0;
  for (i = 2, x = 0; i < m; ++i) {
    while (x && arr[i] != arr[x + 1]) x = fail[x];
    if (arr[i] == arr[x + 1]) ++x;
    fail[i] = x;
    if (x == n) {
      ans = 1;
      break;
    }
  }
  printf(ans ? "Yes" : "No");
  return 0;
}

### Prompt

Please create a solution in cpp 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 BUF = 1000000;
char buf[BUF], *p1, *p2;
inline char getChar() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, BUF, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  register int f = 0;
  register char c;
  while (!isdigit(c = getChar())) {
  }
  do f = f * 10 + (c ^ 48);
  while (isdigit(c = getChar()));
  return f;
}
const int N = 200003;
const long long yyb = 134874928973;
vector<int> g[N];
long long arr[N];
int fail[N];
int main() {
  int n = read(), m = read(), i, x, y;
  for (i = 1; i <= m; ++i) {
    x = read(), y = read();
    if (x > y) swap(x, y);
    g[x].push_back(y - x), g[y].push_back(x - y + n);
  }
  for (i = 1; i <= n; ++i) {
    sort(g[i].begin(), g[i].end());
    for (auto x : g[i]) arr[i] = (arr[i] * n + x) % yyb;
  }
  for (i = 1; i < n; ++i) arr[i + n] = arr[i];
  m = n << 1;
  bool ans = 0;
  for (i = 2, x = 0; i < m; ++i) {
    while (x && arr[i] != arr[x + 1]) x = fail[x];
    if (arr[i] == arr[x + 1]) ++x;
    fail[i] = x;
    if (x == n) {
      ans = 1;
      break;
    }
  }
  printf(ans ? "Yes" : "No");
  return 0;
}

```