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>
inline int getint() {
  int x = 0, p = 1;
  char c = getchar();
  while (c <= 32) c = getchar();
  if (c == 45) p = -p, c = getchar();
  while (c > 32) x = x * 10 + c - 48, c = getchar();
  return x * p;
}
using namespace std;
const int N = 2e5 + 5;
int cnt, n, m, k, R[N], in[N], l[N], r[N], vis[N], ans[N];
queue<int> qu;
vector<int> G[N], oG[N], id, g[N];
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    pq;
void finit() {
  n = getint();
  m = getint();
  k = getint();
  for (int(i) = 0; (i) < (n); (i)++) R[i] = getint();
  for (int(i) = 0; (i) < (m); (i)++) {
    int x = getint() - 1, y = getint() - 1;
    in[y]++;
    oG[y].push_back(x);
    G[x].push_back(y);
  }
  for (int(i) = 0; (i) < (n); (i)++)
    if (!in[i]) {
      id.push_back(i);
      cnt++;
      qu.push(i);
    }
  while (!qu.empty()) {
    int u = qu.front();
    qu.pop();
    for (int(i) = 0; (i) < (G[u].size()); (i)++) {
      int to = G[u][i];
      in[to]--;
      if (!in[to]) {
        cnt++;
        id.push_back(to);
        qu.push(to);
      }
    }
  }
  if (k > n) {
    printf("%d\n", -1);
    exit(0);
  }
  if (cnt != n) {
    printf("%d\n", -1);
    exit(0);
  }
  for (int i = n - 1; i >= 0; i--) {
    int x = id[i];
    l[x] = 1;
    for (int(j) = 0; (j) < (G[x].size()); (j)++) {
      int to = G[x][j];
      l[x] = max(l[to] + 1, l[x]);
    }
    if (l[x] > k) {
      printf("%d\n", -1);
      exit(0);
    }
    if (R[x] > 0) {
      if (l[x] > R[x]) {
        printf("%d\n", -1);
        exit(0);
      }
      l[x] = R[x];
    }
    g[l[x]].push_back(x);
  }
  for (int(i) = 0; (i) < (n); (i)++) {
    int x = id[i];
    r[x] = k;
    for (int(j) = 0; (j) < (oG[x].size()); (j)++) {
      int to = oG[x][j];
      r[x] = min(r[to] - 1, r[x]);
    }
    if (r[x] <= 0) {
      printf("%d\n", -1);
      exit(0);
    }
    if (R[x] > 0) {
      if (r[x] < R[x]) {
        printf("%d\n", -1);
        exit(0);
      }
      r[x] = R[x];
    }
  }
}
void solve() {
  for (int ii = 1; ii <= k; ii++) {
    for (int i : g[ii]) pq.push(make_pair(r[i], i));
    bool flag = false;
    while (pq.size()) {
      pair<int, int> u = pq.top();
      if (u.first != ii) break;
      ans[u.second] = ii;
      flag = true;
      pq.pop();
    }
    if (!flag) {
      if (pq.empty()) {
        printf("%d\n", -1);
        exit(0);
      }
      pair<int, int> u = pq.top();
      pq.pop();
      ans[u.second] = ii;
    }
  }
  for (int(i) = 0; (i) < (n); (i)++) printf("%d ", ans[i]);
}
int main() {
  finit();
  solve();
  return 0;
}