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;
clock_t __stt;
inline void TStart() { __stt = clock(); }
inline void TReport() {
  printf("\nTaken Time : %.3lf sec\n",
         (double)(clock() - __stt) / CLOCKS_PER_SEC);
}
template <typename T>
T MIN(T a, T b) {
  return a < b ? a : b;
}
template <typename T>
T MAX(T a, T b) {
  return a > b ? a : b;
}
template <typename T>
T ABS(T a) {
  return a > 0 ? a : (-a);
}
template <typename T>
void UMIN(T &a, T b) {
  if (b < a) a = b;
}
template <typename T>
void UMAX(T &a, T b) {
  if (b > a) a = b;
}
int n, m, k, l[200005], r[200005], d[200005], vis[200005];
vector<int> adj[200005], rev[200005], ord;
vector<pair<int, int> > stt[200005];
priority_queue<pair<int, int> > pq;
void dfs(int ver) {
  int i;
  if (vis[ver] == 1) {
    printf("-1\n");
    exit(0);
  } else if (vis[ver] == 2)
    return;
  vis[ver] = 1;
  for (i = 0; i < (int)adj[ver].size(); ++i) {
    dfs(adj[ver][i]);
  }
  vis[ver] = 2;
  ord.push_back(ver);
}
int main() {
  int i, j, u, v;
  scanf("%d%d%d", &n, &m, &k);
  for (i = 0; i < n; ++i) {
    scanf("%d", d + i);
    --d[i];
  }
  for (i = 0; i < m; ++i) {
    scanf("%d%d", &u, &v);
    adj[--u].push_back(--v);
    rev[v].push_back(u);
  }
  for (i = 0; i < n; ++i) {
    if (rev[i].empty()) dfs(i);
  }
  for (i = 0; i < n; ++i) {
    if (!vis[i]) {
      printf("-1\n");
      return 0;
    }
  }
  for (i = 0; i < (int)ord.size(); ++i) {
    u = ord[i];
    l[u] = 0;
    if (d[u] >= 0) l[u] = d[u];
    for (j = 0; j < (int)adj[u].size(); ++j) {
      UMAX(l[u], l[adj[u][j]] + 1);
    }
  }
  for (i = (int)ord.size() - 1; i >= 0; --i) {
    u = ord[i];
    r[u] = k - 1;
    if (d[u] >= 0) r[u] = d[u];
    for (j = 0; j < (int)rev[u].size(); ++j) {
      UMIN(r[u], r[rev[u][j]] - 1);
    }
  }
  for (i = 0; i < n; ++i) {
    if (l[i] > r[i]) {
      printf("-1\n");
      return 0;
    }
    stt[l[i]].push_back(make_pair(-r[i], i));
  }
  for (i = 0; i < k; ++i) {
    for (j = 0; j < (int)stt[i].size(); ++j) {
      pq.push(stt[i][j]);
    }
    pair<int, int> tp = make_pair(1, -1);
    while (!pq.empty() && -tp.first < i) {
      tp = pq.top();
      pq.pop();
    }
    if (-tp.first < i) {
      printf("-1\n");
      return 0;
    }
    d[tp.second] = i;
  }
  for (i = (int)ord.size() - 1; i >= 0; --i) {
    u = ord[i];
    if (d[u] == -1) {
      d[u] = r[u];
      for (j = 0; j < (int)rev[u].size(); ++j) {
        UMIN(d[u], d[rev[u][j]] - 1);
      }
      if (d[u] < l[u]) {
        printf("-1\n");
        return 0;
      }
    }
  }
  for (i = 0; i < n; ++i) printf("%d ", d[i] + 1);
  printf("\n");
  return 0;
}