Today is tuesday, that means there is a dispute in JOHNNY SOLVING team again: they try to understand who is Johnny and who is Solving. That's why guys asked Umnik to help them. Umnik gave guys a connected graph with n vertices without loops and multiedges, such that a degree of any vertex is at least 3, and also he gave a number 1 ≤ k ≤ n. Because Johnny is not too smart, he promised to find a simple path with length at least n/k in the graph. In reply, Solving promised to find k simple by vertices cycles with representatives, such that: 

  * Length of each cycle is at least 3. 
  * Length of each cycle is not divisible by 3. 
  * In each cycle must be a representative - vertex, which belongs only to this cycle among all printed cycles. 



You need to help guys resolve the dispute, for that you need to find a solution for Johnny: a simple path with length at least n/k (n is not necessarily divided by k), or solution for Solving: k cycles that satisfy all the conditions above. If there is no any solution - print -1. 

Input

The first line contains three integers n, m and k (1 ≤ k ≤ n ≤ 2.5 ⋅ 10^5, 1 ≤ m ≤ 5 ⋅ 10^5)

Next m lines describe edges of the graph in format v, u (1 ≤ v, u ≤ n). It's guaranteed that v ≠ u and all m pairs are distinct.

It's guaranteed that a degree of each vertex is at least 3.

Output

Print PATH in the first line, if you solve problem for Johnny. In the second line print the number of vertices in the path c (c ≥ n/k). And in the third line print vertices describing the path in route order.

Print CYCLES in the first line, if you solve problem for Solving. In the following lines describe exactly k cycles in the following format: in the first line print the size of the cycle c (c ≥ 3). In the second line print the cycle in route order. Also, the first vertex in the cycle must be a representative.

Print -1 if there is no any solution. The total amount of printed numbers in the output must be at most 10^6. It's guaranteed, that if exists any solution then there is a correct output satisfies this restriction.

Examples

Input


4 6 2
1 2
1 3
1 4
2 3
2 4
3 4


Output


PATH
4
1 2 3 4 

Input


10 18 2
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
2 3
3 4
2 4
5 6
6 7
5 7
8 9
9 10
8 10


Output


CYCLES
4
4 1 2 3 
4
7 1 5 6 
#include <bits/stdc++.h>
using namespace std;
namespace mine {
long long qread() {
  long long ans = 0, f = 1;
  char c = getchar();
  while (c < '0' or c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c and c <= '9') ans = ans * 10 + c - '0', c = getchar();
  return ans * f;
}
void write(long long num) {
  if (num < 0) putchar('-'), num = -num;
  if (num >= 10) write(num / 10);
  putchar('0' + num % 10);
}
void write1(long long num) {
  write(num);
  putchar(' ');
}
void write2(long long num) {
  write(num);
  putchar('\n');
}
template <typename T>
inline bool chmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool chmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
bool IN(long long x, long long l, long long r) { return l <= x and x <= r; }
void GG() {
  puts("-1");
  exit(0);
}
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const int MOD = 998244353;
int mm(const int x) { return x >= MOD ? x - MOD : x; }
template <typename T>
void add(T &x, const int &y) {
  x = (x + y >= MOD ? x + y - MOD : x + y);
}
long long qpower(long long x, long long e, int mod = MOD) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * x % mod;
    x = x * x % mod;
    e >>= 1;
  }
  return ans;
}
long long invm(long long x) { return qpower(x, MOD - 2); }
const int MM = 1e6 + 10;
long long fac[MM], facinv[MM], Inv[MM];
long long Comb(int n, int m) {
  return n < 0 or m < 0 or n < m
             ? 0
             : fac[n] * facinv[m] % MOD * facinv[n - m] % MOD;
}
void PRE() {
  fac[0] = 1;
  for (int i = (1), I = (MM - 1); i <= I; i++) fac[i] = fac[i - 1] * i % MOD;
  facinv[MM - 1] = invm(fac[MM - 1]);
  for (int i = (MM - 1), I = (1); i >= I; i--)
    facinv[i - 1] = facinv[i] * i % MOD;
  Inv[1] = 1;
  for (int i = (2), I = (MM - 1); i <= I; i++)
    Inv[i] = (MOD - MOD / i) * Inv[MOD % i] % MOD;
}
const int N = 3e5 + 10;
vector<int> to[N];
namespace Tr {
int dep[N], in[N], out[N], dfnid, ff[N][21];
void pre(int x, int fa = 0) {
  in[x] = ++dfnid, dep[x] = dep[fa] + 1;
  ff[x][0] = fa;
  for (int i = (1), I = (20); i <= I; i++) ff[x][i] = ff[ff[x][i - 1]][i - 1];
  for (auto y : to[x])
    if (!dep[y]) pre(y, x);
  out[x] = dfnid;
}
int getlca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = (20), I = (0); i >= I; i--)
    if ((1ll << (i)) <= dep[x] - dep[y]) x = ff[x][i];
  if (x == y) return x;
  for (int i = (20), I = (0); i >= I; i--)
    if (ff[x][i] != ff[y][i]) x = ff[x][i], y = ff[y][i];
  return ff[x][0];
}
int getdis(int x, int y) { return dep[x] + dep[y] - 2 * dep[getlca(x, y)]; }
int G, SUM, mxs[N], siz[N];
bool ban[N];
void getrt(int x, int fa = 0) {
  mxs[x] = 0;
  siz[x] = 1;
  for (auto y : to[x])
    if (y != fa and !ban[y])
      getrt(y, x), siz[x] += siz[y], chmax(mxs[x], siz[y]);
  chmax(mxs[x], SUM - siz[x]);
  if (mxs[x] < mxs[G]) G = x;
}
int Fa[N];
void build_dvt(int x, int fa) {
  ban[x] = 1;
  Fa[x] = fa;
  printf("");
  for (auto y : to[x])
    if (!ban[y]) G = 0, getrt(y), build_dvt(G, x);
}
};  // namespace Tr
using namespace Tr;
void main() {
  int n = qread(), m = qread(), k = qread();
  for (int i = (1), I = (m); i <= I; i++) {
    int x = qread(), y = qread();
    to[x].push_back(y), to[y].push_back(x);
  }
  pre(1);
  for (int x = (1), I = (n); x <= I; x++)
    if (dep[x] >= n / k) {
      printf("PATH\n%d\n", dep[x]);
      for (int now = x; now > 0; now = ff[now][0]) write1(now);
      return;
    }
  puts("CYCLES");
  for (int x = (2), I = (n); x <= I; x++)
    if (in[x] == out[x] and k > 0) {
      k--;
      int A = 0, B = 0;
      for (auto y : to[x])
        if (y != ff[x][0]) {
          if (!A)
            A = y;
          else
            B = y;
        }
      if (dep[A] > dep[B]) swap(A, B);
      if ((dep[x] - dep[A] + 1) % 3) {
        write2(dep[x] - dep[A] + 1);
        for (int now = x; now != ff[A][0]; now = ff[now][0]) write1(now);
      } else if ((dep[x] - dep[B] + 1) % 3) {
        write2(dep[x] - dep[B] + 1);
        for (int now = x; now != ff[B][0]; now = ff[now][0]) write1(now);
      } else {
        write2(dep[B] - dep[A] + 1);
        write1(x);
        for (int now = B; now != ff[A][0]; now = ff[now][0]) write1(now);
      }
      puts("");
    }
}
};  // namespace mine
signed main() {
  srand(time(0));
  mine::main();
  printf("");
}
