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 N = 200001;
int n, m, K, r[N], cnt, du[N], Du[N], head[N], Cnt, Head[N], q[N], L[N], U[N],
    mk[N];
struct xint {
  int to, next;
} e[N * 2], E[N * 2];
void addedge(int x, int y) {
  e[++cnt] = (xint){y, head[x]};
  head[x] = cnt;
  du[y]++;
}
void Addedge(int x, int y) {
  E[++Cnt] = (xint){y, Head[x]};
  Head[x] = Cnt;
  Du[y]++;
}
int h, t;
bool instk[N];
void dfs(int x) {
  mk[x] = 1;
  instk[x] = 1;
  for (int i = head[x]; i; i = e[i].next) {
    int y = e[i].to;
    if (instk[y]) {
      cout << -1 << endl;
      exit(0);
    }
    if (mk[y]) continue;
    dfs(y);
  }
  instk[x] = 0;
}
void check() {
  memset(mk, 0, sizeof mk);
  for (int i = 1; i <= n; ++i)
    if (!mk[i]) dfs(i);
}
void work1() {
  h = 0, t = 0;
  memset(mk, 0, sizeof mk);
  memset(q, 0, sizeof q);
  for (int i = 1; i <= n; ++i) L[i] = 1;
  for (int i = 1; i <= n; ++i)
    if (r[i]) q[++t] = i, L[i] = r[i], mk[i] = 1;
  for (int i = 1; i <= n; ++i)
    if (!du[i] && !r[i]) q[++t] = i, L[i] = 1, mk[i] = 1;
  while (h++ < t) {
    int u = q[h];
    for (int i = head[u]; i; i = e[i].next) {
      int v = e[i].to;
      if (mk[v]) continue;
      du[v]--;
      L[v] = max(L[v], L[u] + 1);
      if (!du[v]) q[++t] = v;
    }
  }
}
void work2() {
  h = 0, t = 0;
  memset(mk, 0, sizeof mk);
  memset(q, 0, sizeof q);
  for (int i = 1; i <= n; ++i) U[i] = K;
  for (int i = 1; i <= n; ++i)
    if (r[i]) q[++t] = i, U[i] = r[i], mk[i] = 1;
  for (int i = 1; i <= n; ++i)
    if (!Du[i] && !r[i]) q[++t] = i, U[i] = K, mk[i] = 1;
  while (h++ < t) {
    int u = q[h];
    for (int i = Head[u]; i; i = E[i].next) {
      int v = E[i].to;
      if (mk[v]) continue;
      Du[v]--;
      U[v] = min(U[v], U[u] - 1);
      if (!Du[v]) q[++t] = v;
    }
  }
}
struct yint {
  int x, y, o;
} sL[N], sU[N];
bool cmpx(yint a, yint b) { return a.x < b.x; }
bool cmpy(yint a, yint b) { return a.y < b.y; }
set<pair<int, int> > s;
int ans[N];
void work3() {
  for (int i = 1; i <= n; ++i)
    if (L[i] > U[i]) {
      cout << -1 << endl;
      exit(0);
    }
  for (int i = 1; i <= n; ++i)
    sL[i].x = sU[i].x = L[i], sL[i].y = sU[i].y = U[i], sL[i].o = sU[i].o = i;
  sort(sL + 1, sL + n + 1, cmpx);
  sort(sU + 1, sU + n + 1, cmpy);
  int cnt1 = 1, cnt2 = 1;
  s.insert(make_pair(0, 0));
  s.insert(make_pair(2e9, 2e9));
  for (int i = 1; i <= K; ++i) {
    while (cnt1 <= n && sL[cnt1].x <= i)
      s.insert(make_pair(sL[cnt1].y, sL[cnt1].o)), cnt1++;
    pair<int, int> S = *s.upper_bound(make_pair(0, 0));
    if (S.second > n) {
      cout << -1 << endl;
      exit(0);
    }
    ans[S.second] = i;
    s.erase(S);
    while (cnt2 <= n && sU[cnt2].y <= i) {
      pair<int, int> S(sU[cnt2].y, sU[cnt2].o);
      if (!ans[sU[cnt2].o]) {
        ans[sU[cnt2].o] = i;
        s.erase(S);
      }
      cnt2++;
    }
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  printf("\n");
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= n; ++i) scanf("%d", &r[i]);
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (r[x] && r[y] && r[x] <= r[y]) {
      cout << -1 << endl;
      exit(0);
    }
    addedge(y, x);
    Addedge(x, y);
  }
  check();
  work1();
  work2();
  work3();
}