Little Artem is a very smart programmer. He knows many different difficult algorithms. Recently he has mastered in 2-SAT one.

In computer science, 2-satisfiability (abbreviated as 2-SAT) is the special case of the problem of determining whether a conjunction (logical AND) of disjunctions (logical OR) have a solution, in which all disjunctions consist of no more than two arguments (variables). For the purpose of this problem we consider only 2-SAT formulas where each disjunction consists of exactly two arguments.

Consider the following 2-SAT problem as an example: <image>. Note that there might be negations in 2-SAT formula (like for x1 and for x4).

Artem now tries to solve as many problems with 2-SAT as possible. He found a very interesting one, which he can not solve yet. Of course, he asks you to help him. 

The problem is: given two 2-SAT formulas f and g, determine whether their sets of possible solutions are the same. Otherwise, find any variables assignment x such that f(x) ≠ g(x). 

Input

The first line of the input contains three integers n, m1 and m2 (1 ≤ n ≤ 1000, 1 ≤ m1, m2 ≤ n2) — the number of variables, the number of disjunctions in the first formula and the number of disjunctions in the second formula, respectively.

Next m1 lines contains the description of 2-SAT formula f. The description consists of exactly m1 pairs of integers xi ( - n ≤ xi ≤ n, xi ≠ 0) each on separate line, where xi > 0 corresponds to the variable without negation, while xi < 0 corresponds to the variable with negation. Each pair gives a single disjunction. Next m2 lines contains formula g in the similar format.

Output

If both formulas share the same set of solutions, output a single word "SIMILAR" (without quotes). Otherwise output exactly n integers xi (<image>) — any set of values x such that f(x) ≠ g(x).

Examples

Input

2 1 1
1 2
1 2


Output

SIMILAR


Input

2 1 1
1 2
1 -2


Output

0 0 

Note

First sample has two equal formulas, so they are similar by definition.

In second sample if we compute first function with x1 = 0 and x2 = 0 we get the result 0, because <image>. But the second formula is 1, because <image>.
#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int n, ans[N];
struct SAT {
  inline int get(int x) { return x > 0 ? x * 2 - 2 : -x * 2 - 1; }
  int m;
  bitset<N> a[N];
  void read() {
    int t1, t2;
    for (int i = 1; i <= m; i++) {
      cin >> t1 >> t2;
      t1 = get(t1);
      t2 = get(t2);
      a[t1 ^ 1].set(t2);
      a[t2 ^ 1].set(t1);
    }
  }
  void floyd() {
    for (int i = 0; i < n * 2; i++)
      for (int j = 0; j < n * 2; j++)
        if (a[j][i]) a[j] |= a[i];
  }
  void solve() {
    floyd();
    for (int i = 0; i < n * 2; i++)
      if (a[i][i ^ 1])
        for (int j = 0; j < n * 2; j++) a[i].set(j), a[j].set(i ^ 1);
    floyd();
  }
  void dfs(int pos) {
    if (ans[pos] != -1) return;
    ans[pos] = 1;
    ans[pos ^ 1] = 0;
    for (int i = 0; i < n * 2; i++)
      if (a[pos][i]) ans[i] = 1, ans[i ^ 1] = 0;
  }
  void calc(int x, int y) {
    dfs(x);
    dfs(y ^ 1);
    for (int i = 0; i < n * 2; i++)
      if (ans[i] == -1 && !a[i][i ^ 1]) dfs(i);
  }
} f, g;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> f.m >> g.m;
  f.read();
  f.solve();
  g.read();
  g.solve();
  memset(ans, -1, sizeof(ans));
  for (int i = 0; i < n * 2; i++)
    for (int j = 0; j < n * 2; j++)
      if (i != j && f.a[i][j] != g.a[i][j]) {
        if (f.a[i][j])
          g.calc(i, j);
        else
          f.calc(i, j);
        for (int k = 0; k < n; k++) cout << ans[k * 2] << ' ';
        cout << endl;
        return 0;
      }
  cout << "SIMILAR" << endl;
  return 0;
}
