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#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (b); i >= (a); i--)
#define SZ(x) ((int)x.size())
#define all(x) x.begin(), x.end()
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define st first
#define nd second
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
auto &operator<<(auto &o, pair<auto, auto> p) {
    return o << "(" << p.st << ", " << p.nd << ")";
}
auto operator<<(auto &o, auto x)->decltype(end(x), o) {
    o << "{"; int i=0; for(auto e: x) o << ", " + 2*!i++ << e;
    return o << "}";
}
#define deb(x...) cerr << "[" #x "]: ", [](auto...$) { ((cerr<<$<<"; "),...) << endl; }(x)

const int nax = 1e6 + 5;
const int mod = 1e9 + 7;
vector<int> adj[nax];
int n, m;
int a[nax];
int color[nax];
bool vis[nax];
bool bipartite;
bool nei;
int cnt[2][2];

ll pp(ll a, ll b){
    ll ans = 1;
    while(b){
        if(b & 1){
            ans *= a;
            ans %= mod;
        }
        a *= a;
        a %= mod;
        b /= 2;
    }
    return ans;
}

ll inv(ll a){
    return pp(a, mod - 2);
}

int f[nax];
int iw[nax];

void prep(){
    f[0] = iw[0] = 1;
    rep(i, 1, nax - 1){
        f[i] = (1LL * f[i - 1] * i % mod);
        iw[i] = inv(f[i]);
    }
}

ll binom(ll n, ll k){
    ll ans = f[n];
    ans *= iw[k];
    ans %= mod;
    ans *= iw[n - k];
    ans %= mod;
    return ans;
}

vector<int> now;

void dfs(int v, int p){
    now.pb(v);
    vis[v] = true;
    cnt[p][a[v]] += 1;
    color[v] = p;
    for(int x : adj[v]){
        if(color[x] == color[v]) bipartite = false;
        if(a[x] == a[v]) nei = true;
    }
    for(int x : adj[v]){
        if(!vis[x]) dfs(x, p ^ 1);
    }
}

void solve(){
    cin >> n >> m;
    rep(i, 1, n) cin >> a[i];
    rep(i, 1, m){
        int x, y; cin >> x >> y;
        adj[x].pb(y);
        adj[y].pb(x);
    }
    rep(i, 1, n) color[i] = 2;
    ll ans = 1;
    rep(i, 1, n){
        if(vis[i]) continue;
        bipartite = true;
        nei = false;
        cnt[0][0] = cnt[0][1] = cnt[1][0] = cnt[1][1] = 0;
        now.clear();
        dfs(i, 0);
        if(!bipartite){
            assert(nei);
            int tot = cnt[0][0] + cnt[0][1] + cnt[1][0] + cnt[1][1];
            int ones = cnt[0][1] + cnt[1][1];
            int zeros = tot - ones;
            ll akt = 0;
            for(int c=ones%2;c<=tot;c+=2){
                akt += binom(tot, c);
                akt %= mod;
            }
            ans *= akt;
            ans %= mod;
            continue;
        }
        if(!nei) continue;
        ll akt = 0;
        int dif = cnt[0][1] - cnt[1][1];
        rep(c, 0, cnt[0][1] + cnt[0][0]){
            int npOnes = c - dif;
            if(npOnes < 0 || npOnes > cnt[1][1] + cnt[1][0]) continue;
            akt += 1LL * binom(cnt[0][1] + cnt[0][0], c) * binom(cnt[1][1] + cnt[1][0], npOnes);
            akt %= mod;
        }
        ans *= akt;
        ans %= mod;
    }
    cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    prep();
    int tt = 1;
    //cin >> tt;
    rep(te, 1, tt) solve();
    return 0;
}