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// Arti1990, II UWr

#include <bits/stdc++.h>

#define forr(i, n)                  for(int i=0; i<n; i++)
#define FOREACH(iter, coll)         for(auto iter = coll.begin(); iter != coll.end(); ++iter)
#define FOREACHR(iter, coll)        for(auto iter = coll.rbegin(); iter != coll.rend(); ++iter)
#define lbound(P,K,FUN)             ({auto SSS=P, PPP = P-1, KKK=(K)+1; while(PPP+1!=KKK) {SSS = (PPP+(KKK-PPP)/2); if(FUN(SSS)) KKK = SSS; else PPP = SSS;} PPP;})
#define testy()                     int _tests; cin>>_tests; FOR(_test, 1, _tests)
#define CLEAR(tab)                  memset(tab, 0, sizeof(tab))
#define CONTAIN(el, coll)           (coll.find(el) != coll.end())
#define FOR(i, a, b)                for(int i=a; i<=b; i++)
#define FORD(i, a, b)               for(int i=a; i>=b; i--)
#define MP                          make_pair
#define PB                          push_back
#define ff                          first
#define ss                          second
#define deb(X)                      X;
#define SIZE(coll) 					((int)coll.size())

#define M 1000000007
#define INF 1000000007LL

using namespace std;

int n, m, a, b;
long long silnia[1000007], odwr[1000007];
long long suma[2], ile[2], dwudzielny;

struct wierzch
{
    int kolor;
    int byl;
    int val;
    vector <int> l;
};
wierzch tab[1000007];
/*
int licz_brut()
{
    vector <int> v;
    FOR(i, 1, n)
        v.PB(tab[i].kolor);
    
    set <vector<int>> s;
    deque <vector<int>> q;
    s.insert(v);
    q.PB(v);
    while(!q.empty())
    {
        v = q.front();
        q.pop_front();

        for(auto el : v)
            cout << el << " ";
        cout << '\n';

        FOR(i, 1, n)
            for(auto el : tab[i].l)
            {
                if(el > i)
                {
                    if(v[i-1] == v[el-1])
                    {
                        v[i-1] ^= 1;
                        v[el-1] ^= 1;
                        if(s.count(v) == 0)
                        {
                            s.insert(v);
                            q.PB(v);
                        }
                        v[i-1] ^= 1;
                        v[el-1] ^= 1;
                    }
                }
            }
    }
    return s.size();
}
*/
long long pot(int n)
{
    long long res = 1;
    forr(i, n)
        res = (res*2)%M;
    return res;
}

pair <long long, long long> euklid(long long a, long long b)
{
    if(b == 0)   return make_pair(1, 0);
    pair <long long, long long> p = euklid(b, a%b);
    return make_pair(p.second, p.first - p.second * (a/b));
}

long long odwrotnosc(long long a, long long mod)
{
    long long e = euklid(a, mod).first;
    return (e < 0 ? mod-((-e) % mod) : (e % mod));
}

long long binom(int n, int k)
{
    return silnia[n] * odwr[k] % M * odwr[n-k] % M;
}

void init()
{
    silnia[0] = 1;
    odwr[0] = 1;
    FOR(i, 1, 200007)
    {
        silnia[i] = (silnia[i-1]*i) % M;
        odwr[i] = odwrotnosc(silnia[i], M);
    }
}

void  dfs(int nr, int kolor)
{
    //cout << "odwiedza " << nr << " z kolorem " << kolor << '\n';
    tab[nr].byl = 1;
    ile[kolor]++;
    tab[nr].kolor = kolor;
    suma[kolor] += tab[nr].val;
    for(auto el : tab[nr].l)
        if(!tab[el].byl)
        {
            dfs(el, 1-kolor);
        }
        else if(tab[el].kolor == kolor)
            dwudzielny = 0;
}

long long licz()
{
    long long res = 1;
    FOR(i, 1, n)
        if(!tab[i].byl)
        {
            ile[0] = ile[1] = 0;
            suma[0] = suma[1] = 0;
            dwudzielny = 1;
            dfs(i, 0);
            if(!dwudzielny)
            {
                //cout << "dodaje potege " << ile[0] + ile[1] - 1 << '\n';
                res = (res*pot(ile[0]+ile[1]-1)) % M;
            }
            else
            {
                long long r = 0;
                int minn = min(suma[0], suma[1]);
                suma[0] -= minn;
                suma[1] -= minn;
                int dl = min(ile[0] - suma[0], ile[1] - suma[1]);
                FOR(d, 0, dl)
                {
                    //cout << "dodaje kombinacje " << suma[0]+d << " z " << ile[0] << " razy " << suma[1]+d << " z " << ile[1] << '\n';
                    r = (r + binom(ile[0], suma[0]+d) * binom(ile[1], suma[1]+d)) % M;
                }
                res = res * r % M;
            }
            //cout << "res: " << res << '\n';
        }
    return res;
}

int solve()
{
    init();

    cin >> n >> m;
    FOR(i, 1, n)
        cin >> tab[i].val;
    forr(i, m)
    {
        cin >> a >> b;
        tab[a].l.PB(b);
        tab[b].l.PB(a);
    }

    cout << licz() << '\n';

    //cout << licz_brut() << '\n';

    return 0;
}

int main()
{
    //std::ios_base::sync_with_stdio(0);
    //cin.tie(0);
    //cout.tie(0);

    solve();

    return 0;
}