CodeForces 164A Variable, or There and Back Again（搜索）

题意：

给你n和m，表示有n个状态和m条单向边

快乐路径表示从1开始，2结束的路径，这个路径中间没有1就可以

问你这些状态哪些是快乐路径上的，哪些不是

思路：两次BFS

#include<bits/stdc++.h>using namespace std;const int maxn = 1e6+10;vector<int>e1[maxn];vector<int>e2[maxn];int vis1[maxn];int vis2[maxn];int a[maxn];queue<int>q;int main(){    int n,m;scanf("%d%d",&n,&m);for (int i = 1;i<=n;i++)scanf("%d",&a[i]);for (int i = 1;i<=m;i++){int u,v;scanf("%d%d",&u,&v);e1[u].push_back(v);e2[v].push_back(u);}for (int i = 1;i<=n;i++)if (a[i]==1){vis1[i]=1;q.push(i);}while (!q.empty()){int now = q.front();q.pop();for (int i = 0;i<e1[now].size();i++){int v = e1[now][i];if (a[v]==1)continue;if (vis1[v])continue;vis1[v]=1;q.push(v);}}while (!q.empty())q.pop();for (int i = 1;i<=n;i++)    if (a[i]==2){vis2[i]=1;q.push(i);}while (!q.empty()){int now = q.front();q.pop();for (int i = 0;i<e2[now].size();i++){int v = e2[now][i];if (a[v]==1)vis2[v]=1;;if (vis2[v])continue;vis2[v]=1;q.push(v);}}for (int i = 1;i<=n;i++)if (vis1[i]&&vis2[i])printf("1\n");elseprintf("0\n");}

Description

Life is not easy for the perfectly common variable named Vasya. Wherever it goes, it is either assigned a value, or simply ignored, or is being used!

Vasya's life goes in states of a program. In each state, Vasya can either be used (for example, to calculate the value of another variable), or be assigned a value, or ignored. Between some states are directed (oriented) transitions.

A path is a sequence of states v₁, v₂, ..., v_x, where for any 1 ≤ i < x exists a transition from v_i to v_i + 1.

Vasya's value in state v is interesting to the world, if exists path p₁, p₂, ..., p_k such, that p_i = v for some i(1 ≤ i ≤ k), in state p₁Vasya gets assigned a value, in state p_k Vasya is used and there is no state p_i (except for p₁) where Vasya gets assigned a value.

Help Vasya, find the states in which Vasya's value is interesting to the world.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 10⁵) — the numbers of states and transitions, correspondingly.

The second line contains space-separated n integers f₁, f₂, ..., f_n (0 ≤ f_i ≤ 2), f_i described actions performed upon Vasya in state i: 0represents ignoring, 1 — assigning a value, 2 — using.

Next m lines contain space-separated pairs of integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i), each pair represents the transition from the state number a_i to the state number b_i. Between two states can be any number of transitions.

Output

Print n integers r₁, r₂, ..., r_n, separated by spaces or new lines. Number r_i should equal 1, if Vasya's value in state i is interesting to the world and otherwise, it should equal 0. The states are numbered from 1 to n in the order, in which they are described in the input.

Sample Input

Input

4 31 0 0 21 22 33 4

Output

1111

Input

3 11 0 21 3

Output

101

Input

3 12 0 11 3

Output

000

Hint

In the first sample the program states can be used to make the only path in which the value of Vasya interests the world, 1  2  3  4; it includes all the states, so in all of them Vasya's value is interesting to the world.

The second sample the only path in which Vasya's value is interesting to the world is , — 1  3; state 2 is not included there.

In the third sample we cannot make from the states any path in which the value of Vasya would be interesting to the world, so the value of Vasya is never interesting to the world.