#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
inline char gc() {
  return getchar();
  static const int maxs = 1 << 16;
  static char buf[maxs], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, maxs, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long rd() {
  long long x = 0;
  char c = gc();
  bool neg = 0;
  while (c < '0' || c > '9') {
    if (c == '-') neg = 1;
    c = gc();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = gc();
  return neg ? (~x + 1) : x;
}
int N, M;
pair<int, int> s[maxn];
int num[maxn], cnt, mi[maxn];
inline int getdis(pair<int, int> x) {
  return min((x.second - x.first + N) % N, (x.first - x.second + N) % N);
}
inline int getbeg(pair<int, int> x) {
  return ((x.second - x.first + N) % N == getdis(x)) ? x.first : x.second;
}
inline bool cmp(pair<int, int> a, pair<int, int> b) {
  int la = getdis(a), lb = getdis(b);
  return la == lb ? (getbeg(a) < getbeg(b)) : la < lb;
}
inline bool ju2dge(int x) {
  for (int i = 1; i <= x; i++) {
    for (int j = 0; j < cnt / x - 1; j++)
      if (num[j * x + i] != num[(j + 1) * x + i]) return 0;
  }
  return 1;
}
inline int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
inline long long lcm(int a, int b) { return 1ll * a * b / gcd(a, b); }
bool hav[maxn];
inline bool judge(int x) {
  for (int i = 1, j = 1; i <= M; i = j + 1) {
    for (j = i; j <= M && getdis(s[j + 1]) == getdis(s[j]); j++) {
      hav[getbeg(s[j])] = 1;
    }
    hav[getbeg(s[j])] = 1;
    for (j = i; j <= M && getdis(s[j + 1]) == getdis(s[j]); j++) {
      if (2 * x == N && getdis(s[j]) * 2 == N) continue;
      if (!hav[(getbeg(s[j]) + x - 1) % N + 1]) return 0;
    }
    if (!(2 * x == N && getdis(s[j]) * 2 == N) &&
        !hav[(getbeg(s[j]) + x - 1) % N + 1])
      return 0;
    for (j = i; j <= M && getdis(s[j + 1]) == getdis(s[j]); j++) {
      hav[getbeg(s[j])] = 0;
    }
    hav[getbeg(s[j])] = 0;
  }
  return 1;
}
int main() {
  N = rd(), M = rd();
  for (int i = 1; i <= M; i++) {
    s[i].first = rd(), s[i].second = rd();
  }
  sort(s + 1, s + M + 1, cmp);
  bool bl = 0;
  for (int i = 1; i * i <= N; i++) {
    if (N % i == 0) {
      if (judge(i)) {
        puts("Yes");
        return 0;
      }
      if (i != 1 && judge(N / i)) {
        puts("Yes");
        return 0;
      }
    }
  }
  puts("No");
  return 0;
  for (int i = 1, j = 1; i <= M; i = j + 1) {
    cnt = 0;
    for (j = i; j <= M && getdis(s[j + 1]) == getdis(s[j]); j++) {
      num[++cnt] = getbeg(s[j + 1]) - getbeg(s[j]);
    }
  }
  long long ans = 1;
  for (int i = 1; i <= M; i++) {
    if (!mi[i]) continue;
    ans = lcm(ans, mi[i]) % N;
    if (!ans) {
      bl = 0;
      break;
    }
  }
  if (bl)
    puts("Yes");
  else
    puts("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 maxn = 2e5 + 10;
inline char gc() {
  return getchar();
  static const int maxs = 1 << 16;
  static char buf[maxs], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, maxs, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long rd() {
  long long x = 0;
  char c = gc();
  bool neg = 0;
  while (c < '0' || c > '9') {
    if (c == '-') neg = 1;
    c = gc();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = gc();
  return neg ? (~x + 1) : x;
}
int N, M;
pair<int, int> s[maxn];
int num[maxn], cnt, mi[maxn];
inline int getdis(pair<int, int> x) {
  return min((x.second - x.first + N) % N, (x.first - x.second + N) % N);
}
inline int getbeg(pair<int, int> x) {
  return ((x.second - x.first + N) % N == getdis(x)) ? x.first : x.second;
}
inline bool cmp(pair<int, int> a, pair<int, int> b) {
  int la = getdis(a), lb = getdis(b);
  return la == lb ? (getbeg(a) < getbeg(b)) : la < lb;
}
inline bool ju2dge(int x) {
  for (int i = 1; i <= x; i++) {
    for (int j = 0; j < cnt / x - 1; j++)
      if (num[j * x + i] != num[(j + 1) * x + i]) return 0;
  }
  return 1;
}
inline int gcd(int a, int b) { return a ? gcd(b % a, a) : b; }
inline long long lcm(int a, int b) { return 1ll * a * b / gcd(a, b); }
bool hav[maxn];
inline bool judge(int x) {
  for (int i = 1, j = 1; i <= M; i = j + 1) {
    for (j = i; j <= M && getdis(s[j + 1]) == getdis(s[j]); j++) {
      hav[getbeg(s[j])] = 1;
    }
    hav[getbeg(s[j])] = 1;
    for (j = i; j <= M && getdis(s[j + 1]) == getdis(s[j]); j++) {
      if (2 * x == N && getdis(s[j]) * 2 == N) continue;
      if (!hav[(getbeg(s[j]) + x - 1) % N + 1]) return 0;
    }
    if (!(2 * x == N && getdis(s[j]) * 2 == N) &&
        !hav[(getbeg(s[j]) + x - 1) % N + 1])
      return 0;
    for (j = i; j <= M && getdis(s[j + 1]) == getdis(s[j]); j++) {
      hav[getbeg(s[j])] = 0;
    }
    hav[getbeg(s[j])] = 0;
  }
  return 1;
}
int main() {
  N = rd(), M = rd();
  for (int i = 1; i <= M; i++) {
    s[i].first = rd(), s[i].second = rd();
  }
  sort(s + 1, s + M + 1, cmp);
  bool bl = 0;
  for (int i = 1; i * i <= N; i++) {
    if (N % i == 0) {
      if (judge(i)) {
        puts("Yes");
        return 0;
      }
      if (i != 1 && judge(N / i)) {
        puts("Yes");
        return 0;
      }
    }
  }
  puts("No");
  return 0;
  for (int i = 1, j = 1; i <= M; i = j + 1) {
    cnt = 0;
    for (j = i; j <= M && getdis(s[j + 1]) == getdis(s[j]); j++) {
      num[++cnt] = getbeg(s[j + 1]) - getbeg(s[j]);
    }
  }
  long long ans = 1;
  for (int i = 1; i <= M; i++) {
    if (!mi[i]) continue;
    ans = lcm(ans, mi[i]) % N;
    if (!ans) {
      bl = 0;
      break;
    }
  }
  if (bl)
    puts("Yes");
  else
    puts("No");
  return 0;
}

```