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[200003];
vector<int> vr[200003];
struct ura {
  int pz, vmin, vmax;
};
int rzf[200003];
int fix[200003];
ura vc[200003];
int fv[200003];
int viz[200003], k;
int hasfixed[200003];
void dfsi(int pz) {
  viz[pz] = 1;
  int mxc = 0;
  for (int i = 0; i < v[pz].size(); ++i) {
    int nn = v[pz][i];
    if (viz[nn] && vc[nn].vmin == 0) {
      cout << -1;
      exit(0);
    }
    if (!viz[nn]) dfsi(nn);
    mxc = max(mxc, vc[nn].vmin);
  }
  if (fix[pz])
    vc[pz].vmin = fix[pz];
  else
    vc[pz].vmin = 1 + mxc;
}
void dfsup(int pz) {
  viz[pz] = 1;
  int mnc = k + 1;
  for (int i = 0; i < vr[pz].size(); ++i) {
    int nn = vr[pz][i];
    if (!viz[nn]) dfsup(nn);
    mnc = min(mnc, vc[nn].vmax);
  }
  if (fix[pz])
    vc[pz].vmax = fix[pz];
  else
    vc[pz].vmax = mnc - 1;
}
bool cmp(ura a, ura b) { return a.vmin < b.vmin; }
int elemx;
int heap[200003];
bool cmp1(int a, int b) { return vc[a].vmax < vc[b].vmax; }
void insert(int val) {
  heap[++elemx] = val;
  int pz = elemx;
  while (pz > 1 && cmp1(heap[pz], heap[pz >> 1])) {
    swap(heap[pz], heap[pz >> 1]);
    pz >>= 1;
  }
}
void del() {
  swap(heap[1], heap[elemx]);
  --elemx;
  int pz = 1;
  while ((pz << 1) <= elemx) {
    int fiu = (pz << 1);
    if (fiu + 1 <= elemx && cmp1(heap[fiu + 1], heap[fiu])) ++fiu;
    if (cmp1(heap[fiu], heap[pz])) {
      swap(heap[pz], heap[fiu]);
      pz = fiu;
    } else
      break;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int t, i, j, n, m;
  cin >> n >> m >> k;
  for (i = 1; i <= n; ++i) {
    cin >> fix[i];
    vc[i].pz = i;
  }
  for (i = 1; i <= m; ++i) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    vr[b].push_back(a);
  }
  for (i = 1; i <= n; ++i)
    if (!viz[i]) dfsi(i);
  for (i = 1; i <= n; ++i)
    if (vc[i].vmin > k) {
      cout << -1;
      return 0;
    }
  for (i = 1; i <= n; ++i) viz[i] = 0;
  for (i = 1; i <= n; ++i)
    if (!viz[i]) dfsup(i);
  for (i = 1; i <= n; ++i)
    if (vc[i].vmax < 1 || vc[i].vmax < vc[i].vmin) {
      cout << -1;
      return 0;
    }
  sort(vc + 1, vc + n + 1, cmp);
  j = 1;
  for (int kc = 1; kc <= k; ++kc) {
    while (j <= n && vc[j].vmin == kc) insert(j++);
    int ok = 0;
    while (elemx > 0 && vc[heap[1]].vmax == kc) {
      ok = 1;
      rzf[vc[heap[1]].pz] = kc;
      del();
    }
    if (ok) continue;
    if (elemx == 0) {
      cout << -1;
      return 0;
    }
    rzf[vc[heap[1]].pz] = kc;
    del();
  }
  while (elemx > 0) {
    rzf[vc[heap[1]].pz] = k;
    del();
  }
  int ok = 1;
  for (i = 1; i <= n; ++i) ++fv[rzf[i]];
  for (i = 1; i <= n; ++i) {
    for (j = 0; j < v[i].size(); ++i) {
      if (rzf[i] <= rzf[v[i][j]]) ok = 0;
    }
  }
  for (i = 1; i <= k; ++i)
    if (!fv[i]) ok = 0;
  if (!ok) {
    cout << -1;
    return 0;
  }
  for (i = 1; i <= n; ++i) cout << rzf[i] << ' ';
}