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, m1, m2;
struct LBC {
  int n, m, nn, sol, c[N];
  bitset<N> e[N];
  void color(int x) {
    if (c[x] != -1) return;
    c[x] = 1;
    c[x ^ 1] = 0;
    for (int i = 0; i < nn; i++)
      if (e[x][i]) color(i);
  }
  void init(int _n, int _m) {
    n = _n;
    m = _m;
    nn = n << 1;
    for (int i = 1; i <= m; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      x = (x > 0 ? (x - 1) * 2 + 1 : (-x - 1) * 2);
      y = (y > 0 ? (y - 1) * 2 + 1 : (-y - 1) * 2);
      e[x ^ 1][y] = e[y ^ 1][x] = 1;
    }
    for (int i = 0; i < nn; i++) e[i][i] = 1;
    for (int k = 0; k < nn; k++)
      for (int i = 0; i < nn; i++)
        if (e[i][k]) e[i] |= e[k];
    sol = 1;
    for (int i = 0; i < nn; i++)
      if (e[i][i ^ 1] && e[i ^ 1][i]) sol = 0;
    if (sol) {
      for (int i = 0; i < nn; i++) c[i] = -1;
      for (int i = 0; i < nn; i++)
        if (e[i][i ^ 1]) color(i ^ 1);
    }
  }
  void solve(int c1 = -1, int c2 = -1) {
    if (~c1) color(c1);
    if (~c2) color(c2);
    for (int i = 0; i < nn; i++)
      if (c[i] == -1) color(i);
    for (int i = 0; i < n; i++) putchar(1 - c[i << 1] + '0'), putchar(' ');
  }
} f, g;
bool solve(LBC &f, LBC &g) {
  if (!f.sol) return 0;
  if (!g.sol) return f.solve(), 1;
  for (int i = 0; i < n * 2; i++)
    if (f.c[i] && !g.c[i]) return f.solve(i), 1;
  for (int i = 0; i < n * 2; i++)
    if (f.c[i] == -1)
      for (int j = i; j < n * 2; j++)
        if (f.c[j] == -1)
          if (!f.e[i][j] && g.e[i][j]) return f.solve(i, j ^ 1), 1;
  return 0;
}
int main() {
  scanf("%d%d%d", &n, &m1, &m2);
  f.init(n, m1);
  g.init(n, m2);
  if (!solve(f, g) && !solve(g, f)) puts("SIMILAR");
}
