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 = 2222;
bitset<N> a[2][N];
int x, y, m[2], ok[N], n;
int transform(int x) { return (x > 0) ? ((x - 1) * 2) : ((-x - 1) * 2 + 1); }
void build_it(int m, bitset<N>* a) {
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &x, &y);
    x = transform(x), y = transform(y);
    a[x ^ 1].set(y);
    a[y ^ 1].set(x);
  }
  for (int i = 0; i < 2 * n; ++i) a[i].set(i);
  for (int k = 0; k < 2 * n; ++k)
    for (int i = 0; i < 2 * n; ++i)
      if (a[i][k]) a[i] |= a[k];
  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 k = 0; k < 2 * n; ++k)
    for (int i = 0; i < 2 * n; ++i)
      if (a[i][k]) a[i] |= a[k];
}
void markiza(bitset<N>* a, int x) {
  if (ok[x] != -1) return;
  ok[x] = 1, ok[x ^ 1] = 0;
  for (int i = 0; i < 2 * n; ++i)
    if (a[x][i]) markiza(a, i);
}
void restore(bitset<N>* a, int xtrue, int xfalse) {
  markiza(a, xtrue);
  markiza(a, xfalse ^ 1);
  for (int i = 0; i < 2 * n; ++i)
    if ((ok[i] == -1) && !a[i][i ^ 1]) {
      markiza(a, i);
    }
}
int main() {
  scanf("%d%d%d", &n, &m[0], &m[1]);
  for (int j = 0; j < 2; ++j) build_it(m[j], a[j]);
  for (int i = 0; i < 2 * n; ++i) ok[i] = -1;
  for (int i = 0; i < 2 * n; ++i)
    for (int j = 0; j < 2 * n; ++j)
      if (a[0][i][j] != a[1][i][j]) {
        restore(a[a[1][i][j] ? 0 : 1], i, j);
        for (int k = 0; k < n; ++k)
          printf("%d%c", ok[2 * k], " \n"[k == n - 1]);
        return 0;
      }
  printf("SIMILAR\n");
  return 0;
}
