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>
#pragma GCC optimize "-O3"
using namespace std;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
inline int readChar();
template <class T = int>
inline T readInt();
template <class T>
inline void writeInt(T x, char end = 0);
inline void writeChar(int x);
inline void writeWord(const char *s);
static const int buf_size = 4096;
inline int getChar() {
  static char buf[buf_size];
  static int len = 0, pos = 0;
  if (pos == len) {
    pos = 0, len = fread(buf, 1, buf_size, stdin);
  }
  if (pos == len) {
    return -1;
  }
  return buf[pos++];
}
inline int readChar() {
  int c = getChar();
  while (c <= 32) {
    c = getChar();
  }
  return c;
}
template <class T>
inline T readInt() {
  int s = 1, c = readChar();
  T x = 0;
  if (c == '-') s = -1, c = getChar();
  while ('0' <= c && c <= '9') x = x * 10 + c - '0', c = getChar();
  return s == 1 ? x : -x;
}
static int write_pos = 0;
static char write_buf[buf_size];
inline void writeChar(int x) {
  if (write_pos == buf_size)
    fwrite(write_buf, 1, buf_size, stdout), write_pos = 0;
  write_buf[write_pos++] = x;
}
template <class T>
inline void writeInt(T x, char end) {
  if (x < 0) writeChar('-'), x = -x;
  char s[24];
  int n = 0;
  while (x || !n) s[n++] = '0' + x % 10, x /= 10;
  while (n--) writeChar(s[n]);
  if (end) writeChar(end);
}
inline void writeWord(const char *s) {
  while (*s) writeChar(*s++);
}
struct Flusher {
  ~Flusher() {
    if (write_pos) fwrite(write_buf, 1, write_pos, stdout), write_pos = 0;
  }
} flusher;
const string FILENAME = "input";
const int MAXN = 3e5 + 1;
int n, m, k;
vector<int> edge[MAXN];
bool used[MAXN];
int minc[MAXN], maxc[MAXN], c[MAXN], nc[MAXN];
vector<int> top_sort;
vector<int> event[MAXN];
set<pair<int, int>> res;
set<int> need;
void dfs(int i) {
  used[i] = true;
  for (int j : edge[i]) {
    if (!used[j]) {
      dfs(j);
    }
  }
  top_sort.push_back(i);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
    --c[i];
    nc[i] = c[i];
    if (c[i] != -1) need.insert(c[i]);
    maxc[i] = k - 1;
  }
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> b >> a;
    edge[a - 1].push_back(b - 1);
  }
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      dfs(i);
    }
  }
  reverse(top_sort.begin(), top_sort.end());
  for (int i : top_sort) {
    if (c[i] == -1) {
      nc[i] = min(minc[i], k - 1);
    }
    used[i] = false;
    for (int j : edge[i]) {
      chkmax(minc[j], nc[i] + 1);
    }
  }
  reverse(top_sort.begin(), top_sort.end());
  for (int i : top_sort) {
    if (c[i] != -1) {
      maxc[i] = c[i];
    } else {
      for (int j : edge[i]) {
        chkmin(maxc[i], maxc[j] - 1);
      }
      chkmax(maxc[i], 0);
      event[maxc[i]].push_back(i);
    }
  }
  for (int id = k; id--;) {
    for (auto i : event[id]) res.insert({nc[i], i});
    if (need.count(id)) continue;
    while (res.size()) {
      auto t = *res.rbegin();
      res.erase(t);
      if (minc[t.second] <= id && id <= maxc[t.second]) {
        c[t.second] = id;
        need.insert(id);
        break;
      } else {
        c[t.second] = t.first;
        need.insert(t.first);
      }
    }
  }
  while (res.size()) {
    auto t = *res.rbegin();
    res.erase(t);
    c[t.second] = t.first;
  }
  for (int i = 0; i < n; ++i) {
    for (int j : edge[i]) {
      if (c[i] >= c[j]) {
        cout << -1 << endl;
        exit(0);
      }
    }
  }
  if (need.size() != k) {
    cout << -1 << endl;
    exit(0);
  }
  for (int i = 0; i < n; ++i) {
    cout << c[i] + 1 << ' ';
  }
}