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;
int n;
int m, k;
struct edge {
  int x, y;
} ed[300000];
bool flag;
vector<int> head[300000];
int high[300000], low[300000], du[300000], q[300000], ans[300000], top;
vector<pair<int, int>> jx[300000];
set<pair<int, int>> S;
void topsort(int now[], int delta) {
  int l = 0, r = -1, v, to;
  for (int i = 1; i <= n; ++i)
    if (du[i] == 0) q[++r] = i;
  while (l <= r) {
    v = q[l++];
    for (int i = 0; i < head[v].size(); ++i) {
      to = head[v][i];
      if (delta < 0)
        now[to] = min(now[to], now[v] + delta);
      else
        now[to] = max(now[to], now[v] + delta);
      if (--du[to] == 0) q[++r] = to;
    }
  }
  if (r != n - 1) {
    flag = false;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    high[i] = k;
    low[i] = 1;
  }
  int x;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &x);
    if (x != 0) high[i] = low[i] = x;
  }
  flag = true;
  for (int i = 1; i <= m; ++i) scanf("%d%d", &ed[i].x, &ed[i].y);
  for (int i = 1; i <= n; ++i) head[i].clear(), du[i] = 0;
  for (int i = 1; i <= m; ++i) head[ed[i].x].push_back(ed[i].y), ++du[ed[i].y];
  topsort(high, -1);
  for (int i = 1; i <= n; ++i) head[i].clear(), du[i] = 0;
  for (int i = 1; i <= m; ++i) head[ed[i].y].push_back(ed[i].x), ++du[ed[i].x];
  topsort(low, 1);
  if (!flag) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    if (low[i] > high[i]) {
      printf("-1\n");
      return 0;
    }
  top = k;
  for (int i = 1; i <= k; ++i) jx[i].clear();
  S.clear();
  for (int i = 1; i <= n; ++i) jx[high[i]].push_back(make_pair(-low[i], i));
  pair<int, int> temp;
  set<pair<int, int>>::iterator it;
  for (int i = 0; i < jx[k].size(); ++i) S.insert(jx[k][i]);
  while (top) {
    if (S.empty()) {
      printf("-1\n");
      return 0;
    }
    it = S.begin();
    temp = *it;
    ans[temp.second] = top;
    S.erase(it);
    while (1) {
      it = S.begin();
      temp = *it;
      if (-temp.first < top) break;
      ans[temp.second] = top;
      S.erase(it);
    }
    --top;
    for (int i = 0; i < jx[top].size(); ++i) S.insert(jx[top][i]);
  }
  while (!S.empty()) {
    it = S.begin();
    temp = *it;
    ans[temp.second] = 1;
    S.erase(it);
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}