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 = 2010;
int n;
int ans[N];
void watch(bitset<N> a) {
  string s = a.to_string();
  reverse(s.begin(), s.end());
  cout << s << endl;
}
struct _2sat {
  bitset<N> g[N];
  int m;
  void debug() {
    for (int i = 0; i < 2 * n; i++) watch(g[i]);
  }
  int get(int x) {
    if (x > 0)
      return x * 2 - 2;
    else
      return -x * 2 - 1;
  }
  void read() {
    while (m--) {
      int x, y;
      scanf("%d%d", &x, &y);
      x = get(x), y = get(y);
      g[x ^ 1].set(y);
      g[y ^ 1].set(x);
    }
  }
  void calc() {
    for (int i = 0; i < 2 * n; i++)
      for (int j = 0; j < 2 * n; j++)
        if (g[j][i]) g[j] |= g[i];
  }
  void solve() {
    calc();
    for (int i = 0; i < n * 2; i++)
      if (g[i][i ^ 1]) {
        for (int j = 0; j < n * 2; j++) {
          g[i].set(j);
          g[j].set(i ^ 1);
        }
      }
    calc();
  }
  void dfs(int v) {
    if (ans[v] != -1) return;
    ans[v] = 1, ans[v ^ 1] = 0;
    for (int i = 0; i < n * 2; i++)
      if (g[v][i]) ans[i] = 1, ans[i ^ 1] = 0;
  }
  void getans(int u, int v) {
    dfs(u);
    dfs(v ^ 1);
    for (int i = 0; i < n * 2; i++)
      if (ans[i] == -1 && !g[i][i ^ 1]) dfs(i);
  }
} F, G;
int main() {
  scanf("%d%d%d", &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.g[i][j] != G.g[i][j]) {
        if (F.g[i][j])
          G.getans(i, j);
        else
          F.getans(i, j);
        for (int k = 0; k < n; k++) printf("%d ", ans[k * 2]);
        puts("");
        return 0;
      }
    }
  }
  puts("SIMILAR");
}
