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;
vector<int> v[200020], w[200020];
int r[200020], mn[200020], deg[200020], x, y, ans[200020], id[200020];
int n, m, k;
queue<int> q;
bitset<200020> vis;
bool cmp(int x, int y) { return mn[x] > mn[y]; }
struct cmp1 {
  bool operator()(const int &x, const int &y) const { return mn[x] < mn[y]; }
};
priority_queue<int, vector<int>, cmp1> q1, q2;
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", r + i);
    vis[r[i]] = true;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    w[x].push_back(y);
    v[y].push_back(x);
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    deg[i] = w[i].size();
    if (deg[i] == 0) {
      q.push(i);
      cnt++;
    }
  }
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (auto y : v[cur]) {
      deg[y]--;
      if (deg[y] == 0) {
        q.push(y);
        cnt++;
      }
    }
    int now = 0;
    for (auto y : w[cur]) {
      if (now < mn[y]) now = mn[y];
    }
    now++;
    if (now > k) {
      puts("-1");
      return 0;
    }
    if (r[cur] && now > r[cur]) {
      puts("-1");
      return 0;
    }
    if (r[cur] == 0)
      mn[cur] = now;
    else
      mn[cur] = r[cur];
  }
  if (cnt < n) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    deg[i] = v[i].size();
    if (!deg[i]) {
      if (r[i])
        q2.push(i);
      else
        q1.push(i);
    }
  }
  int idx, pos = k;
  while (!q1.empty() || !q2.empty()) {
    while (vis[pos] && pos > 0) {
      pos--;
    }
    if (q2.empty()) {
      idx = q1.top();
      q1.pop();
    } else {
      int idd = q2.top();
      if (r[idd] > pos || q1.empty()) {
        idx = idd;
        q2.pop();
      } else {
        idx = q1.top();
        q1.pop();
      }
    }
    for (auto y : w[idx]) {
      deg[y]--;
      if (!deg[y]) {
        if (r[y])
          q2.push(y);
        else
          q1.push(y);
      }
    }
    if (r[idx] && r[idx] < pos) {
      puts("-1");
      return 0;
    }
    if (mn[idx] > pos) {
      ans[idx] = mn[idx];
      continue;
    }
    int now = k + 1;
    for (auto y : v[idx]) {
      if (now > ans[y]) now = ans[y];
    }
    now--;
    if (now < pos) {
      puts("-1");
      return 0;
    } else {
      ans[idx] = pos;
      vis[pos] = true;
    }
  }
  while (vis[pos] && pos > 0) {
    pos--;
  }
  if (pos) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
}