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;
int T, n, m[2], p, k;
int ans[2010];
struct graph {
  bitset<2010> a[2010];
  void solve() {
    for (int i = 0; i < 2 * n; i++)
      for (int j = 0; j < 2 * n; j++)
        if (a[j][i]) a[j] |= a[i];
    for (int i = 0; i < 2 * n; i++)
      if (a[i][i ^ 1])
        for (int j = 0; j < 2 * n; j++) {
          a[i].set(j);
          a[j].set(i ^ 1);
        }
    for (int i = 0; i < 2 * n; i++)
      for (int j = 0; j < 2 * n; j++)
        if (a[j][i]) a[j] |= a[i];
  }
  void work(int u) {
    if (ans[u] >= 0) return;
    ans[u] = true, ans[u ^ 1] = false;
    for (int i = 0; i < 2 * n; i++)
      if (a[u][i]) ans[i] = true, ans[i ^ 1] = false;
  }
  void doit(int x, int y) {
    work(x);
    work(y ^ 1);
    for (int i = 0; i < 2 * n; i++)
      if (ans[i] < 0 && !a[i][i ^ 1]) work(i);
  }
} f[2];
int main() {
  scanf("%d%d%d", &n, &m[0], &m[1]);
  for (int k = 0; k <= 1; k++) {
    for (int i = 1, u, v; i <= m[k]; i++) {
      scanf("%d%d", &u, &v);
      u = (abs(u) - 1) << 1 | (u > 0);
      v = (abs(v) - 1) << 1 | (v > 0);
      f[k].a[u ^ 1].set(v);
      f[k].a[v ^ 1].set(u);
    }
    f[k].solve();
  }
  for (int i = 0; i < 2 * n; i++) ans[i] = -1;
  for (int i = 0; i < 2 * n; i++)
    for (int j = 0; j < 2 * n; j++)
      if (i != j && f[0].a[i][j] != f[1].a[i][j]) {
        if (f[0].a[i][j]) f[1].doit(i, j);
        if (f[1].a[i][j]) f[0].doit(i, j);
        for (int i = 0; i < n; i++) printf("%d ", ans[i << 1 | 1]);
        puts("");
        return 0;
      }
  puts("SIMILAR");
}
