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 maxn = 2222, mod = 998244353;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 0;
  while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
inline int qmo(int x) { return x + (x >> 31 ? mod : 0); }
int n, m1, m2;
struct twosat {
  bitset<maxn> e[maxn];
  int in[maxn];
  inline void add(int u, int v) { e[u][v] = true; }
  void floyd() {
    for (int i = (2); i <= (2 * n + 1); i++) e[i][i] = true;
    for (int k = (2); k <= (2 * n + 1); k++)
      for (int i = (2); i <= (2 * n + 1); i++)
        if (e[i][k]) e[i] |= e[k];
  }
  void work() {
    floyd();
    for (int i = (2); i <= (2 * n + 1); i++)
      if (e[i][i ^ 1])
        for (int j = (2); j <= (2 * n + 1); j++) add(i, j), add(j, i ^ 1);
    floyd();
  }
  void dfs(int u) {
    if (in[u]) return;
    in[u] = 1;
    in[u ^ 1] = -1;
    for (int v = (2); v <= (2 * n + 1); v++)
      if (e[u][v]) dfs(v);
  }
  void solve(int x, int y) {
    dfs(x);
    dfs(y);
    for (int i = (1); i <= (n); i++)
      if (!in[2 * i]) {
        if (e[2 * i][2 * i + 1])
          dfs(2 * i + 1);
        else
          dfs(2 * i);
      }
    for (int i = (1); i <= (n); i++) printf("%d ", in[2 * i] == 1 ? 0 : 1);
  }
} f, g;
int main() {
  n = read();
  m1 = read();
  m2 = read();
  while (m1--) {
    int u = read(), v = read();
    u = u > 0 ? 2 * u : -2 * u + 1;
    v = v > 0 ? 2 * v : -2 * v + 1;
    f.add(u, v ^ 1);
    f.add(v, u ^ 1);
  }
  f.work();
  while (m2--) {
    int u = read(), v = read();
    u = u > 0 ? 2 * u : -2 * u + 1;
    v = v > 0 ? 2 * v : -2 * v + 1;
    g.add(u, v ^ 1);
    g.add(v, u ^ 1);
  }
  g.work();
  for (int i = (2); i <= (2 * n + 1); i++)
    for (int j = (2); j <= (2 * n + 1); j++)
      if (f.e[i][j] ^ g.e[i][j] && i != j) {
        if (f.e[i][j])
          return g.solve(i, j ^ 1), 0;
        else
          return f.solve(i, j ^ 1), 0;
      }
  puts("SIMILAR");
}
