#include <bits/stdc++.h>
using namespace std;
typedef struct {
  int a;
  int b;
  int d;
  int p;
} S;
int c, i, j, k, l, m, n, t, u, y, q[999], z[100005];
S w[400005];
int V(S x, S y) { return x.a < y.a || x.a == y.a && x.b < y.b; }
int W(S x, S y) { return x.d < y.d || x.d == y.d && x.a < y.a; }
int main() {
  for (scanf("%d%d", &n, &m), l = m; i < m; i++) {
    scanf("%d%d", &w[i].a, &w[i].b);
    w[i].d = w[i].b - w[i].a + (w[i].b > w[i].a ? 0 : n);
    w[i].a > w[i].b ? t = w[i].a, w[i].a = w[i].b, w[i].b = t,
                      w[i].d = n - w[i].d : 0;
    w[i].d* 2 > n ? t = w[i].a, w[i].a = w[i].b, w[i].b = t,
                    w[i].d = n - w[i].d : 0;
    w[i].b += w[i].b < w[i].a ? n : 0;
    if (w[i].d * 2 == n)
      w[l].a = w[i].b, w[l].b = w[i].a + n, w[l].d = w[i].d, l++;
  }
  for (i = 0, m = l; i++ < n;) n % i ?: q[c++] = i;
  sort(w, w + m, W);
  for (i = 0, k = m, t = 1; i < m; i++) {
    if (w[i].d == w[i + 1].d)
      t++;
    else {
      t > k ?: k = t;
      for (u = i - t + 1; u <= i; u++) w[u].p = t;
      t = 1;
    }
  }
  sort(w, w + m, V);
  for (; q[--c] > k;)
    ;
  for (; c; c--) {
    t = m / q[c], u = n / q[c];
    if (m % q[c] < 1 && n % q[c] < 1) {
      for (i = 0; i < m - t; i++) {
        if (w[i].a + u ^ w[i + t].a || w[i].b + u ^ w[i + t].b) goto G;
      }
      y = q[c];
    }
  G:;
  }
  puts(y ? "Yes" : "No");
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
typedef struct {
  int a;
  int b;
  int d;
  int p;
} S;
int c, i, j, k, l, m, n, t, u, y, q[999], z[100005];
S w[400005];
int V(S x, S y) { return x.a < y.a || x.a == y.a && x.b < y.b; }
int W(S x, S y) { return x.d < y.d || x.d == y.d && x.a < y.a; }
int main() {
  for (scanf("%d%d", &n, &m), l = m; i < m; i++) {
    scanf("%d%d", &w[i].a, &w[i].b);
    w[i].d = w[i].b - w[i].a + (w[i].b > w[i].a ? 0 : n);
    w[i].a > w[i].b ? t = w[i].a, w[i].a = w[i].b, w[i].b = t,
                      w[i].d = n - w[i].d : 0;
    w[i].d* 2 > n ? t = w[i].a, w[i].a = w[i].b, w[i].b = t,
                    w[i].d = n - w[i].d : 0;
    w[i].b += w[i].b < w[i].a ? n : 0;
    if (w[i].d * 2 == n)
      w[l].a = w[i].b, w[l].b = w[i].a + n, w[l].d = w[i].d, l++;
  }
  for (i = 0, m = l; i++ < n;) n % i ?: q[c++] = i;
  sort(w, w + m, W);
  for (i = 0, k = m, t = 1; i < m; i++) {
    if (w[i].d == w[i + 1].d)
      t++;
    else {
      t > k ?: k = t;
      for (u = i - t + 1; u <= i; u++) w[u].p = t;
      t = 1;
    }
  }
  sort(w, w + m, V);
  for (; q[--c] > k;)
    ;
  for (; c; c--) {
    t = m / q[c], u = n / q[c];
    if (m % q[c] < 1 && n % q[c] < 1) {
      for (i = 0; i < m - t; i++) {
        if (w[i].a + u ^ w[i + t].a || w[i].b + u ^ w[i + t].b) goto G;
      }
      y = q[c];
    }
  G:;
  }
  puts(y ? "Yes" : "No");
}

```