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 maxn = 2e5 + 10;
vector<int> nex[maxn], pre[maxn];
int in[maxn], out[maxn];
int down[maxn], up[maxn];
int rk[maxn], n, m, k;
int num[maxn];
struct node {
  int v, id;
  node() {}
  node(int _v, int _id) {
    v = _v;
    id = _id;
  }
  bool operator<(const node &t) const {
    return (v == t.v) ? up[id] < up[t.id] : v > t.v;
  }
} p[maxn];
queue<int> Q;
priority_queue<pair<int, int> > P;
bool bfs() {
  while (!Q.empty()) Q.pop();
  for (int i = 1; i <= n; i++) {
    if (out[i] == 0) Q.push(i);
    down[i] = (rk[i] == 0) ? 1 : rk[i];
  }
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    for (int i = 0; i < pre[v].size(); i++) {
      int u = pre[v][i];
      if (rk[u] && down[u] > rk[u]) return false;
      down[u] = max(down[u], down[v] + 1);
      if (--out[u] == 0) Q.push(u);
    }
  }
  for (int i = 1; i <= n; i++)
    if (out[i] || down[i] > k) return false;
  return true;
}
int UP(int v) {
  int ans = k + 1;
  for (int i = 0; i < pre[v].size(); i++) ans = min(down[pre[v][i]] - 1, ans);
  return ans;
}
bool d[maxn];
void dell(int u) {
  if (d[u]) return;
  d[u] = true;
  for (int i = 0; i < nex[u].size(); i++) {
    int v = nex[u][i];
    in[v]--;
    if (in[v] == 0 && rk[v] == 0) P.push(make_pair(down[v], v));
  }
}
int main() {
  int u, v;
  while (~scanf("%d%d%d", &n, &m, &k)) {
    for (int i = 1; i <= n; i++) {
      nex[i].clear();
      pre[i].clear();
      in[i] = out[i] = 0;
    }
    for (int i = 1; i <= n; i++) scanf("%d", &rk[i]);
    bool flag = true;
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &u, &v);
      if (rk[u] > 0 && rk[v] > 0 && rk[v] >= rk[u]) {
        flag = false;
        continue;
      }
      nex[u].push_back(v);
      out[u]++;
      pre[v].push_back(u);
      in[v]++;
    }
    if (!flag) {
      puts("-1");
      continue;
    }
    if (!bfs()) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= k; i++) num[i] = 0;
    while (!P.empty()) P.pop();
    for (int i = 1; i <= n; i++) {
      num[down[i]]++;
      up[i] = UP(i);
      p[i] = node(down[i], i);
      if (in[i] == 0 && rk[i] == 0) P.push(make_pair(down[i], i));
      ;
    }
    sort(p + 1, p + 1 + n);
    for (int i = 1; i <= n; i++) d[i] = false;
    int pos = 1;
    flag = true;
    for (int i = k; i >= 1; i--)
      if (num[i] == 0) {
        while (pos <= n && p[pos].v > i) dell(p[pos++].id);
        while (!P.empty() && P.top().first > i) {
          dell(P.top().second);
          P.pop();
        }
        if (P.empty()) {
          flag = false;
          break;
        }
        num[P.top().first]--;
        down[P.top().second] = i;
        dell(P.top().second);
        P.pop();
      }
    if (!flag)
      puts("-1");
    else {
      for (int i = 1; i < n; i++) printf("%d ", down[i]);
      printf("%d\n", down[n]);
    }
  }
  return 0;
}