Variable, or There and Back Again

 CodeForces - 164A 

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_ksuch, that p_i = v for some i (1 ≤ i ≤ k), in state p₁ Vasya 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_idescribed actions performed upon Vasya in state i: 0 represents 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.

Examples Input

4 3
1 0 0 2
1 2
2 3
3 4

Output

1
1
1
1

Input

3 1
1 0 2
1 3

Output

1
0
1

Input

3 1
2 0 1
1 3

Output

0
0
0

Note

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.

 

题意：好像就是从所有1走到2,的路径覆盖的点答案是1，否则答案是0

sol：很容易发现就是减一下反图，分别从1和2开始bfs

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0; bool f=0; char ch=' ';
    while(!isdigit(ch))    {f|=(ch=='-'); ch=getchar();}
    while(isdigit(ch)) {s=(s<<3)+(s<<1)+(ch^48); ch=getchar();}
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0) {putchar('-'); x=-x;}
    if(x<10) {putchar(x+'0'); return;}
    write(x/10); putchar((x%10)+'0');
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=100005,M=200005;
int n,m,a[N];
int tot=0,Next[M],to[M],head[N];
bool arr1[N],arr2[N];
inline void Link(int x,int y)
{
    Next[++tot]=head[x]; to[tot]=y; head[x]=tot;
}
#define PB push_back
vector<int>E[N];
int main()
{
    queue<int>Que;
    int i,e,x,y;
    R(n); R(m);
    for(i=1;i<=n;i++) R(a[i]);
    for(i=1;i<=m;i++)
    {
        R(x); R(y); Link(x,y); E[y].PB(x);
    }
    memset(arr1,0,sizeof arr1);
    for(i=1;i<=n;i++) if(a[i]==1) {Que.push(i); arr1[i]=1;}
    while(!Que.empty())
    {
        x=Que.front(); Que.pop();
        for(e=head[x];e;e=Next[e])
        {
            if(arr1[to[e]]) continue;
            arr1[to[e]]=1; Que.push(to[e]);
        }
    }
    while(!Que.empty()) Que.pop();
    memset(arr2,0,sizeof arr2);
    for(i=1;i<=n;i++) if(a[i]==2) {Que.push(i); arr2[i]=1;}
    while(!Que.empty())
    {
        x=Que.front(); Que.pop();
        for(i=0;i<E[x].size();i++)
        {
            if(a[E[x][i]]==1) {arr2[E[x][i]]=1; continue;}
            if(arr2[E[x][i]]) continue;
            arr2[E[x][i]]=1; Que.push(E[x][i]);
        }
    }
    for(i=1;i<=n;i++) {if(arr1[i]&&arr2[i]) puts("1"); else puts("0");}
    return 0;
}

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