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 INF = 1 << 30;
const long long inf = 1LL << 60;
const int N = 200010;
struct Edge {
  int to, next;
} e1[N], e2[N];
int list1[N], list2[N], d1, d2;
priority_queue<pair<int, int> > Q;
queue<int> S;
bool vis[N];
vector<int> v[N];
void add(int x, int y, int *list, int &d, Edge *e) {
  e[d].to = y, e[d].next = list[x], list[x] = d++;
}
int n, m, k;
int ans[N], Min[N], din[N], dout[N], d[N];
int read() {
  int w(0), f(0);
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') f = 1, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return f ? -w : w;
}
void Imps() {
  puts("-1");
  exit(0);
}
void Delete(int x) {
  int y;
  for (int i = list2[x]; i >= 0; i = e2[i].next) {
    y = e2[i].to;
    d[y]--;
    if (!d[y] && !ans[y]) Q.push(make_pair(Min[y], y));
  }
}
int main() {
  n = read(), m = read(), k = read();
  memset(list1, -1, 4 * (n + 1)), memset(list2, -1, 4 * (n + 1));
  for (int i = 1; i <= n; ++i)
    ans[i] = read(), vis[ans[i]] = 1, v[ans[i]].push_back(i);
  int x, y;
  pair<int, int> tmp;
  for (int i = 1; i <= m; ++i) {
    x = read(), y = read();
    add(y, x, list1, d1, e1);
    add(x, y, list2, d2, e2);
    din[x]++, dout[y]++;
  }
  for (int i = 1; i <= n; ++i) {
    d[i] = din[i];
    if (!d[i]) S.push(i);
  }
  while (!S.empty()) {
    x = S.front(), S.pop();
    if (ans[x]) {
      if (ans[x] < Min[x])
        Imps();
      else
        Min[x] = ans[x];
    } else if (!Min[x])
      Min[x] = 1;
    for (int i = list1[x]; i >= 0; i = e1[i].next) {
      y = e1[i].to;
      Min[y] = max(Min[y], Min[x] + 1);
      d[y]--;
      if (!d[y]) S.push(y);
    }
  }
  for (int i = 1; i <= n; ++i) {
    d[i] = dout[i];
    if (!d[i] && !ans[i]) Q.push(make_pair(Min[i], i));
  }
  for (int i = k; i >= 1; --i) {
    for (int j = 0; j < v[i].size(); ++j) {
      x = v[i][j];
      Delete(x);
    }
    while (!Q.empty()) {
      tmp = Q.top();
      if (tmp.first > i)
        Imps();
      else if (tmp.first == i)
        ans[tmp.second] = i, vis[i] = 1, Q.pop(), Delete(tmp.second);
      else
        break;
    }
    if (!vis[i]) {
      if (!Q.empty()) {
        tmp = Q.top();
        ans[tmp.second] = i, vis[i] = 1, Q.pop(), Delete(tmp.second);
      } else
        Imps();
    }
  }
  for (x = 1; x <= n; ++x)
    for (int i = list1[x]; i >= 0; i = e1[i].next)
      if (ans[x] >= ans[e1[i].to]) Imps();
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}