Kitty is a little cat. She is crazy about a game recently.

There arenscenes in the game(mark from 1 ton). Each scene has a numberp_i. Kitty's score will become least_common_multiple(x,p_i) when Kitty enter thei^thscene.xis the score that Kitty had previous. Notice that Kitty will become mad If she go to another scene but the score didn't change.

Kitty is staying in the first scene now(withp₁score). Please find out how many paths which can arrive at then^thscene and haskscores at there. Of course, you can't make Kitty mad.

We regard two paths different if and only if the edge sequence is different.

There are multiple test cases. For each test case:

The first line contains three integern(2 ≤n≤ 2000),m(2 ≤m≤ 20000),k(2 ≤k≤ 10⁶). Then followed bymlines. Each line contains two integeru,v(1 ≤u,v≤ n, u ≠ v) indicate we can go tov^thscene fromu^thscene directly. The last line of each case contains n integerp_i(1 ≤p_i≤ 10⁶).

Process to the end of input.

One line for each case. The number of paths module 1000000007.

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 #include <map>

 using namespace std;

 typedef long long LL;

 const int MAXN = ;

 const int MAXE = ;

 const int MOD = ;

 map<int, int> mp;

 int n, m, k, ecnt;

 int head[MAXN];

 int to[MAXE], next[MAXE];

 int dp[MAXN][MAXN], val[MAXN];

 void init() {

     memset(head, , sizeof(head));

     ecnt = ;

 }

 inline void add_edge(int u, int v) {

     to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++;

 }

 LL gcd(LL a, LL b) {

     return b ? gcd(b, a % b) : a;

 }

 LL lcm(LL a, LL b) {

     return a * b / gcd(a, b);

 }

 inline void get_app() {

     mp.clear();

     int cnt = ;

     for(int i = ; i * i <= k; ++i) {

         if(k % i != ) continue;

         mp[i] = ++cnt;

         if(i * i != k) mp[k / i] = ++cnt;

     }

 }

 int dfs(int u, LL x) {

     if(u == n && x == k) return ;

     int now = mp[x];

     if(dp[u][now] != -) return dp[u][now];

     dp[u][now] = ;

     for(int p = head[u]; p; p = next[p]) {

         int &v = to[p];

         if(k % val[v] != ) continue;

         LL tmp = lcm(x, val[v]);

         if(tmp == x || tmp > k) continue;

         dp[u][now] = (dp[u][now] + dfs(v, tmp)) % MOD;

     }

     return dp[u][now];

 }

 int main() {

     while(scanf("%d%d%d", &n, &m, &k) != EOF) {

         init();

         while(m--) {

             int u, v;

             scanf("%d%d", &u, &v);

             add_edge(u, v);

         }

         for(int i = ; i <= n; ++i) scanf("%d", &val[i]);

         get_app();

         memset(dp, , sizeof(dp));

         int ans = (k % val[]) ?  : dfs(, val[]);

         cout<<ans<<endl;

     }

 }