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 = 200005;
set<pair<int, int> > S1, S2;
vector<int> first[maxn], G[maxn];
int s[maxn], ss[maxn], a[maxn];
bool p[maxn];
queue<int> Q;
void work(int x) {
  for (int i = 0; i < (int)G[x].size(); ++i) {
    int y = G[x][i];
    --ss[y];
    if (ss[y] == 0) {
      if (!p[y])
        S1.insert(pair<int, int>(-a[y], y));
      else
        S2.insert(pair<int, int>(-a[y], y));
    }
  }
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] != 0) p[i] = true;
  }
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    first[y].push_back(x);
    ++s[x];
    ++ss[y];
  }
  for (int i = 1; i <= n; ++i) {
    if (s[i] == 0) Q.push(i);
  }
  bool flag = true;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    int mm = 0;
    for (int i = 0; i < (int)G[x].size(); ++i) {
      int y = G[x][i];
      mm = max(mm, a[y]);
    }
    if (a[x] != 0 && a[x] <= mm) flag = false;
    if (a[x] == 0) a[x] = mm + 1;
    if (a[x] > k) flag = false;
    for (int i = 0; i < (int)first[x].size(); ++i) {
      int y = first[x][i];
      --s[y];
      if (s[y] == 0) {
        Q.push(y);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (ss[i] == 0) {
      if (!p[i])
        S1.insert(pair<int, int>(-a[i], i));
      else
        S2.insert(pair<int, int>(-a[i], i));
    }
  }
  for (int i = k; i >= 1; --i) {
    bool first = false;
    while (!S2.empty() && S2.begin()->first <= -i) {
      first = true;
      int x = S2.begin()->second;
      S2.erase(S2.begin());
      work(x);
    }
    while (!S1.empty() && S1.begin()->first <= -i) {
      first = true;
      int x = S1.begin()->second;
      S1.erase(S1.begin());
      work(x);
    }
    if (!first) {
      if (S1.empty()) {
        flag = false;
        break;
      }
      int x = S1.begin()->second;
      S1.erase(S1.begin());
      a[x] = i;
      work(x);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < (int)G[i].size(); ++i) {
      if (a[i] <= a[G[i][j]]) flag = false;
    }
  }
  if (!flag)
    puts("-1");
  else {
    for (int i = 1; i <= n; ++i) {
      printf("%d", a[i]);
      if (i == n)
        puts("");
      else
        printf(" ");
    }
  }
}