Problem: There are n military men in the Berland army. Some of them have given orders to other military men by now. Given m pairs (xi, yi), meaning that the military man xi gave the i-th order to another military man yi.

It is time for reform! The Berland Ministry of Defence plans to introduce ranks in the Berland army. Each military man should be assigned a rank — integer number between 1 and k, inclusive. Some of them have been already assigned a rank, but the rest of them should get a rank soon.

Help the ministry to assign ranks to the rest of the army so that:

  * for each of m orders it is true that the rank of a person giving the order (military man xi) is strictly greater than the rank of a person receiving the order (military man yi); 
  * for each rank from 1 to k there is at least one military man with this rank. 

Input

The first line contains three integers n, m and k (1 ≤ n ≤ 2·105, 0 ≤ m ≤ 2·105, 1 ≤ k ≤ 2·105) — number of military men in the Berland army, number of orders and number of ranks.

The second line contains n integers r1, r2, ..., rn, where ri > 0 (in this case 1 ≤ ri ≤ k) means that the i-th military man has been already assigned the rank ri; ri = 0 means the i-th military man doesn't have a rank yet.

The following m lines contain orders one per line. Each order is described with a line containing two integers xi, yi (1 ≤ xi, yi ≤ n, xi ≠ yi). This line means that the i-th order was given by the military man xi to the military man yi. For each pair (x, y) of military men there could be several orders from x to y.

Output

Print n integers, where the i-th number is the rank of the i-th military man. If there are many solutions, print any of them.

If there is no solution, print the only number -1.

Examples

Input

5 3 3
0 3 0 0 2
2 4
3 4
3 5


Output

1 3 3 2 2 


Input

7 6 5
0 4 5 4 1 0 0
6 1
3 6
3 1
7 5
7 1
7 4


Output

2 4 5 4 1 3 5 


Input

2 2 2
2 1
1 2
2 1


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 500010;
int n, m, K, i, a[N], f[N], x, y, g[N], v[N], nxt[N], ed, d[N];
int h, t, q[N];
int G[N], V[N], NXT[N], ED;
set<int> T;
int cnt[N];
struct E {
  int x, y;
} e[N];
priority_queue<pair<int, int> > qu;
inline void add(int x, int y) {
  v[++ed] = y;
  nxt[ed] = g[x];
  g[x] = ed;
  d[y]++;
}
inline void ADD(int x, int y) {
  V[++ED] = y;
  NXT[ED] = G[x];
  G[x] = ED;
  d[y]++;
}
bool can_vio() {
  long long t = 1;
  for (int i = 1; i <= n; i++)
    if (!a[i]) {
      t *= K;
      if (t > 1e9) return 0;
    }
  return 1;
}
inline bool check() {
  int i, j;
  for (i = 1; i <= n; i++) {
    for (int j = g[i]; j; j = nxt[j])
      if (f[i] >= f[v[j]]) return 0;
  }
  static bool used[N];
  for (i = 1; i <= K; i++) used[i] = 0;
  for (i = 1; i <= n; i++) used[f[i]] = 1;
  for (i = 1; i <= K; i++)
    if (!used[i]) return 0;
  return 1;
}
void dfs(int x) {
  if (x > n) {
    if (check()) {
      for (int i = 1; i <= n; i++) printf("%d ", f[i]);
      exit(0);
    }
    return;
  }
  if (a[x]) {
    f[x] = a[x];
    dfs(x + 1);
    return;
  }
  for (int i = 1; i <= K; i++) {
    f[x] = i;
    dfs(x + 1);
  }
}
int main() {
  srand(time(NULL));
  scanf("%d%d%d", &n, &m, &K);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), f[i] = max(1, a[i]);
  for (i = 1; i <= m; i++) scanf("%d%d", &e[i].x, &e[i].y);
  random_shuffle(e + 1, e + m + 1);
  for (i = 1; i <= m; i++) {
    x = e[i].x, y = e[i].y;
    add(y, x);
  }
  for (h = i = 1; i <= n; i++)
    if (!d[i]) q[++t] = i;
  while (h <= t) {
    x = q[h++];
    for (i = g[x]; i; i = nxt[i]) {
      f[v[i]] = max(f[v[i]], f[x] + 1);
      if (!(--d[v[i]])) q[++t] = v[i];
    }
  }
  if (t < n) return puts("-1"), 0;
  for (i = 1; i <= n; i++)
    if (a[i] && f[i] != a[i]) return puts("-1"), 0;
  for (i = 1; i <= n; i++)
    if (f[i] > K) return puts("-1"), 0;
  if (can_vio()) {
  }
  for (i = 1; i <= K; i++) T.insert(i);
  for (i = 1; i <= n; i++)
    if (a[i]) T.erase(a[i]);
  for (i = 1; i <= n; i++) T.erase(f[i]);
  for (i = 1; i <= n; i++) cnt[f[i]]++;
  for (i = 1; i <= n; i++) d[i] = 0;
  for (i = 1; i <= m; i++) {
    x = e[i].x, y = e[i].y;
    ADD(x, y);
  }
  for (i = 1; i <= n; i++)
    if (!d[i]) qu.push(pair<int, int>(f[i], i));
  while (!qu.empty()) {
    pair<int, int> t = qu.top();
    qu.pop();
    x = t.second;
    for (i = G[x]; i; i = NXT[i])
      if (!(--d[V[i]])) qu.push(pair<int, int>(f[V[i]], V[i]));
    if (a[x]) continue;
    int lim = K;
    for (int j = g[x]; j; j = nxt[j]) {
      lim = min(lim, f[v[j]] - 1);
    }
    set<int>::iterator it = T.lower_bound(lim);
    if (it != T.begin() && it == T.end()) it--;
    while (it != T.begin() && (*it) > lim) it--;
    bool flag = 0;
    int old = f[x];
    if (it != T.end() && (*it) <= lim) {
      if ((*it) > f[x]) f[x] = *it, flag = 1;
    }
    if (flag) {
      cnt[old]--;
      if (!cnt[old]) T.insert(old);
      cnt[f[x]]++;
      T.erase(f[x]);
    }
  }
  if (T.size()) return puts("-1"), 0;
  for (i = 1; i <= n; i++)
    if (a[i] && f[i] != a[i]) return puts("-1"), 0;
  for (i = 1; i <= n; i++) {
    for (int j = g[i]; j; j = nxt[j])
      if (f[i] >= f[v[j]]) return puts("-1"), 0;
  }
  for (i = 1; i <= n; i++) printf("%d ", f[i]);
}