You are given a simple undirected graph with n vertices, n is even. You are going to write a letter on each vertex. Each letter should be one of the first k letters of the Latin alphabet.

A path in the graph is called Hamiltonian if it visits each vertex exactly once. A string is called palindromic if it reads the same from left to right and from right to left. A path in the graph is called palindromic if the letters on the vertices in it spell a palindromic string without changing the order.

A string of length n is good if: 

  * each letter is one of the first k lowercase Latin letters; 
  * if you write the i-th letter of the string on the i-th vertex of the graph, there will exist a palindromic Hamiltonian path in the graph. 



Note that the path doesn't necesserily go through the vertices in order 1, 2, ..., n.

Count the number of good strings.

Input

The first line contains three integers n, m and k (2 ≤ n ≤ 12; n is even; 0 ≤ m ≤ (n ⋅ (n-1))/(2); 1 ≤ k ≤ 12) — the number of vertices in the graph, the number of edges in the graph and the number of first letters of the Latin alphabet that can be used.

Each of the next m lines contains two integers v and u (1 ≤ v, u ≤ n; v ≠ u) — the edges of the graph. The graph doesn't contain multiple edges and self-loops.

Output

Print a single integer — number of good strings.

Examples

Input


4 3 3
1 2
2 3
3 4


Output


9


Input


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


Output


21


Input


12 19 12
1 3
2 6
3 6
3 7
4 8
8 5
8 7
9 4
5 9
10 1
10 4
10 6
9 10
11 1
5 11
7 11
12 2
12 5
12 11


Output


456165084
#include <bits/stdc++.h>
void sc(int &x) { scanf("%d", &x); }
void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
void sc(long long &x) { scanf("%lld", &x); }
void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
void sc(long long &x, long long &y, long long &z) {
  scanf("%lld%lld%lld", &x, &y, &z);
}
void sc(char *x) { scanf("%s", x); }
void sc(char *x, char *y) { scanf("%s%s", x, y); }
void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
void out(int x) { printf("%d\n", x); }
void out(long long x) { printf("%lld\n", x); }
void out(int x, int y) { printf("%d %d\n", x, y); }
void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
void out(long long x, long long y, long long z) {
  printf("%lld %lld %lld\n", x, y, z);
}
void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
using namespace std;
const int N = 5e5 + 5, mod = 998244353;
int n, m, k;
long long f[15], c[15][15];
bool e[15][15];
long long g[15];
long long ans;
long long st1, st2;
long long C(int n, int m) {
  if (m == 0 || m == n) return 1;
  if (m == 1) return n;
  if (c[n][m]) return c[n][m];
  return c[n][m] = C(n - 1, m - 1) + C(n - 1, m);
}
vector<pair<int, int>> vc;
map<long long, bool> vis;
bool judge(vector<pair<int, int>> &vc) {
  if (vis.count(st1)) return vis[st1];
  int up = (1 << (vc.size())) - 1;
  bool dp[up + 1][vc.size()];
  memset(dp, false, sizeof(dp));
  for (int i = 0; i < vc.size(); i++) dp[1 << i][i] = true;
  for (int i = 0; i < up; i++)
    for (int j = 0; j < vc.size(); j++)
      if (dp[i][j]) {
        for (int k = 0; k < vc.size(); k++)
          if (!(i >> k & 1)) {
            if (e[vc[k].first][vc[j].first] && e[vc[k].second][vc[j].second] ||
                e[vc[k].first][vc[j].second] && e[vc[k].second][vc[j].first])
              dp[i | (1 << k)][k] = true;
          }
      }
  bool ans = false;
  for (int i = 0; i < vc.size(); i++)
    if (e[vc[i].first][vc[i].second]) ans |= dp[up][i];
  return vis[st1] = ans;
}
set<long long> ok;
void cal(vector<pair<int, int>> &vc, int mx) {
  if (judge(vc)) {
    if (!ok.count(st2)) {
      ans += C(k, mx + 1) * f[mx + 1];
      ok.insert(st2);
    }
  }
}
void dfs(int up, int mx) {
  if (up == 0) {
    cal(vc, mx);
    return;
  }
  for (int i = 0; i < n; i++)
    if (up >> i & 1) {
      for (int j = i + 1; j < n; j++)
        if (up >> j & 1) {
          st1 += (i + 1) * g[vc.size() << 1];
          st1 += (j + 1) * g[vc.size() << 1 | 1];
          vc.push_back({i + 1, j + 1});
          int nst = up ^ (1 << i) ^ (1 << j);
          for (int s = 0; s <= mx + 1 && s < k; s++) {
            st2 += (s + 1) * g[i];
            st2 += (s + 1) * g[j];
            dfs(nst, max(s, mx));
            st2 -= (s + 1) * g[i];
            st2 -= (s + 1) * g[j];
          }
          vc.pop_back();
          st1 -= (i + 1) * g[vc.size() << 1];
          st1 -= (j + 1) * g[vc.size() << 1 | 1];
        }
      break;
    }
}
void ways() {
  long long dp[13][13];
  memset(dp, 0, sizeof(dp));
  dp[1][0] = 1;
  for (int i = 1; i < 12; i++)
    for (int j = 0; j < 12; j++)
      for (int k = 0; k <= j + 1 && k < 12; k++)
        dp[i + 1][max(j, k)] += dp[i][j];
  long long ans = 0;
  for (int k = 0; k < 12; k++) ans += dp[12][k];
  out(ans);
}
void sol(int cas) {
  f[0] = g[0] = 1;
  for (int i = 1; i < 15; i++) f[i] = f[i - 1] * i, g[i] = g[i - 1] * 13;
  sc(n, m, k);
  for (int i = 1; i <= m; i++) {
    int x, y;
    sc(x, y);
    e[x][y] = e[y][x] = true;
  }
  int up = (1 << n) - 1;
  dfs(up, -1);
  out(ans);
}
int main() {
  srand(time(0));
  int t = 1, cas = 0;
  while (t--) {
    sol(++cas);
  }
  return 0;
}
