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;
inline int read() {
  int num = 0;
  char c = getchar();
  if (c == '-') return -read();
  while (c >= '0' && c <= '9') {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num;
}
inline long long READ() {
  long long num = 0;
  char c = getchar();
  if (c == '-') return -READ();
  while (c >= '0' && c <= '9') {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num;
}
int l[200007], r[200007];
int deg[200007];
int tps[200007], tppppppppppfxxk;
vector<int> nei[200007], fnei[200007];
bool lgl[200007], lgl2[200007];
vector<int> dsf;
set<pair<int, int> > ugl;
int n, m, k;
int rr[200007];
set<int> cnu;
int n1[200007];
int chs[200007];
queue<int> q;
priority_queue<pair<pair<int, int>, int> > qb;
pair<pair<int, int>, int> bb[200007];
bool FxxK(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.first.second < b.first.second;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) scanf("%d", rr + i);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (!ugl.count(make_pair(a, b))) {
      nei[a].push_back(b);
      fnei[b].push_back(a);
    }
    deg[b]++;
  }
  for (int i = 1; i <= n; i++)
    if (!deg[i]) q.push(i);
  if (!((int)(q).size())) return printf("-1"), 0;
  while (((int)(q).size())) {
    int x = q.front();
    tps[x] = ++tppppppppppfxxk;
    dsf.push_back(x);
    q.pop();
    for (int i = 0; i < ((int)(nei[x]).size()); i++)
      if (!tps[nei[x][i]]) {
        deg[nei[x][i]]--;
        if (!deg[nei[x][i]]) q.push(nei[x][i]);
      }
  }
  if (tppppppppppfxxk < n) return printf("-1"), 0;
  for (int ii = n - 1; ii >= 0; ii--) {
    int i = dsf[ii];
    int maxv = 0;
    for (int j = 0; j < ((int)(nei[i]).size()); j++)
      maxv = max(maxv, l[nei[i][j]]);
    if (rr[i]) {
      if (rr[i] <= maxv) return printf("-1\n"), 0;
      l[i] = rr[i];
    } else
      l[i] = maxv + 1;
  }
  for (int ii = 0; ii < n; ii++) {
    int i = dsf[ii];
    int maxv = k + 1;
    for (int j = 0; j < ((int)(fnei[i]).size()); j++)
      maxv = min(maxv, r[fnei[i][j]]);
    if (rr[i]) {
      if (rr[i] >= maxv) return printf("-1\n"), 0;
      r[i] = rr[i];
    } else
      r[i] = maxv - 1;
  }
  for (int i = 1; i <= n; i++) bb[i] = (make_pair(make_pair(-r[i], l[i]), i));
  sort(bb + 1, bb + n + 1, FxxK);
  int cur = 1;
  for (int i = 1; i <= k; i++) {
    while (cur <= n && bb[cur].first.second == i) {
      qb.push(bb[cur]);
      cur++;
    }
    if (!((int)(qb).size())) return printf("-1\n"), 0;
    bool deled = 0;
    while (((int)(qb).size()) && qb.top().first.first == -i) {
      chs[qb.top().second] = i;
      qb.pop();
      deled = 1;
    }
    if (!deled) {
      if (!((int)(qb).size())) return printf("-1\n"), 0;
      chs[qb.top().second] = i;
      qb.pop();
    }
  }
  for (int i = 1; i <= n; i++)
    if (!chs[i]) return printf("-1"), 0;
  for (int i = 1; i <= n; i++) printf("%d%c", chs[i], i == n ? '\n' : ' ');
  return 0;
}