#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <complex>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <iomanip>
#include <assert.h>
#include <array>
#include <cstdio>
#include <cstring>
#include <random>
#include <functional>
#include <numeric>
#include <bitset>

using namespace std;

#define REP(i,a,b) for(int i=a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define all(c) (c).begin(), (c).end()
#define zero(a) memset(a, 0, sizeof a)
#define minus(a) memset(a, -1, sizeof a)
#define minimize(a, x) a = std::min(a, x)
#define maximize(a, x) a = std::max(a, x)

constexpr int inf = 1<<29;

int const MOD = 1e9+7;
typedef long long ll;


constexpr int MATH_COMBINATION_MAX = 1010;

/***********************************************************************************/
#define DEF_MAX_BOUND(assignment) enum { assignment };
#define MEMO_BEGIN(Identifier)  \
namespace memorized { namespace Identifier {  \
  bool __computed = false; inline bool computed() {if(__computed) { return true; } else { __computed = true; return false; }}
#define MEMO_END }}

#define USE_MEMO(Identifier)        using memorized::Identifier::
#define COMPUTED_RESULT(Identifier) if(memorized::Identifier::computed()) return
#define USE_CONSTANT(Identifier)    using memorized::Identifier::
/***********************************************************************************/


/*
  スターリング数
  異なるN個をちょうどKグループに分けるときの方法の数
  高々Kグループに分ける場合は、グループi(1<=i<=K)をN回選ぶ方法の数となり、これは単なる重複順列でK^Nとなる
  Verified: http://yukicoder.me/problems/663
 */

namespace math { namespace combination {

MEMO_BEGIN(stirling_number)
  DEF_MAX_BOUND(max_value = MATH_COMBINATION_MAX)
  ll dp[max_value][max_value];
MEMO_END

int stirling_number(int N, int K) {
  USE_MEMO(stirling_number)         dp;
  COMPUTED_RESULT(stirling_number)  dp[N][K];

  USE_CONSTANT(stirling_number) max_value;

  REP(i, 1, max_value) {  // undefined when i = 0
    dp[i][1] = dp[i][i] = 1;
    REP(j, 2, i) {
      dp[i][j] = dp[i-1][j-1] + j * dp[i-1][j];
      dp[i][j] %= MOD;
    }
  }

  return dp[N][K];
}

}}

int N, M;
vector<int> g[222];
ll fact[MATH_COMBINATION_MAX];
ll dp[300][300];

int main() {

  cin >> N >> M;

  vector<ll> ws(N);
  rep(i, N) {
    cin >> ws[i];
  }

  rep(i, M) {
    int a, b; cin >> a >> b; a--, b--;
    g[a].push_back(b);
  }

  zero(dp);
  rep(i, N) {
    dp[i][i] = ws[i];
    rep(j, g[i].size()) {
      int tar = g[i][j];
      dp[i][tar] = min(dp[i][i], ws[tar]);
    }
  }

  rep(k, N) rep(i, N) rep(j, N) {
    maximize(dp[i][j], min(dp[i][k], dp[k][j]));
  }

  fact[0] = 1;
  rep(i, MATH_COMBINATION_MAX) fact[i+1] = fact[i] * (i+1) % MOD;

  // 異なるN個を異なるK個に振り分ける方法の数(全射の数) := S(N, K) * K!
  using math::combination::stirling_number;

  ll ans = 0;

  rep(i, N) rep(j, N) {
    if(dp[i][j] < ws[j]) { continue; }
    ans += stirling_number(ws[i], ws[j]) * fact[ws[j]] % MOD;
    ans %= MOD;
  }

  cout << ans << endl;

  return 0;
}