Problem: Phoenix's homeland, the Fire Nation had n cities that were connected by m roads, but the roads were all destroyed by an earthquake. The Fire Nation wishes to repair n-1 of these roads so that all the cities are connected again. 

The i-th city has a_i tons of asphalt. x tons of asphalt are used up when repairing a road, and to repair a road between i and j, cities i and j must have at least x tons of asphalt between them. In other words, if city i had a_i tons of asphalt and city j had a_j tons, there would remain a_i+a_j-x tons after repairing the road between them. Asphalt can be moved between cities if the road between them is already repaired.

Please determine if it is possible to connect all the cities, and if so, output any sequence of roads to repair.

Input

The first line contains integers n, m, and x (2 ≤ n ≤ 3 ⋅ 10^5; n-1 ≤ m ≤ 3 ⋅ 10^5; 1 ≤ x ≤ 10^9) — the number of cities, number of roads, and amount of asphalt needed to repair one road.

The next line contains n space-separated integer a_i (0 ≤ a_i ≤ 10^9) — the amount of asphalt initially at city i.

The next m lines contains two integers x_i and y_i (x_i≠ y_i; 1 ≤ x_i, y_i ≤ n) — the cities connected by the i-th road. It is guaranteed that there is at most one road between each pair of cities, and that the city was originally connected before the earthquake.

Output

If it is not possible to connect all the cities, print NO. Otherwise, print YES followed by n-1 integers e_1, e_2, ..., e_{n-1}, the order in which the roads should be repaired. e_i is the index of the i-th road to repair. If there are multiple solutions, print any.

Examples

Input


5 4 1
0 0 0 4 0
1 2
2 3
3 4
4 5


Output


YES
3
2
1
4


Input


2 1 2
1 1
1 2


Output


YES
1


Input


2 1 2
0 1
1 2


Output


NO


Input


5 6 5
0 9 4 0 10
1 2
1 3
2 3
3 4
1 4
4 5


Output


YES
6
4
1
2

Note

In the first example, the roads are repaired in the following order: 

  * Road 3 is repaired, connecting cities 3 and 4. City 4 originally had 4 tons of asphalt. After this road is constructed, 3 tons remain. 
  * Road 2 is repaired, connecting cities 2 and 3. The asphalt from city 4 can be transported to city 3 and used for the road. 2 tons remain. 
  * Road 1 is repaired, connecting cities 1 and 2. The asphalt is transported to city 2 and used for the road. 1 ton remain. 
  * Road 4 is repaired, connecting cities 4 and 5. The asphalt is transported to city 4 and used for the road. No asphalt remains. 

All the cities are now connected.

In the second example, cities 1 and 2 use all their asphalt together to build the road. They each have 1 ton, so together they have 2 tons, which is enough.

In the third example, there isn't enough asphalt to connect cities 1 and 2.

Solution (in PYTHON3):

Solution:

from bisect import bisect,bisect_left

from collections import *
from heapq import *
from math import gcd,ceil,sqrt,floor,inf

from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *

#------------------------------------------------------------------------
import os
import sys

from io import BytesIO, IOBase
# region fastio
 
BUFSIZE = 8192
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
 

#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def A(n):return [0]*n
def AI(n,x): return [x]*n
def A2(n,m): return [[0]*m for i in range(n)]
def G(n): return [[] for i in range(n)]
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
#------------------------------------------------------------------------


from types import GeneratorType
 
 
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
 
    return wrappedfunc

mod=10**9+7
farr=[1]
ifa=[]
 
def fact(x,mod=0):
    if mod:
        while x>=len(farr):
            farr.append(farr[-1]*len(farr)%mod)
    else:
        while x>=len(farr):
            farr.append(farr[-1]*len(farr))
    return farr[x]
 
def ifact(x,mod):
    global ifa
    fact(x,mod)
    ifa.append(pow(farr[-1],mod-2,mod))
    for i in range(x,0,-1):
        ifa.append(ifa[-1]*i%mod)
    ifa.reverse()
 
def per(i,j,mod=0):
    if i<j: return 0
    if not mod:
        return fact(i)//fact(i-j)
    return farr[i]*ifa[i-j]%mod
    
def com(i,j,mod=0):
    if i<j: return 0
    if not mod:        
        return per(i,j)//fact(j)
    return per(i,j,mod)*ifa[j]%mod
 
def catalan(n):
    return com(2*n,n)//(n+1)
 
def isprime(n):
    for i in range(2,int(n**0.5)+1):
        if n%i==0:
            return False
    return True

def floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
    if a==0:return b//c*(n+1)
    if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
    m=(a*n+b)//c
    return n*m-floorsum(c,c-b-1,a,m-1)

def inverse(a,m):
    a%=m
    if a<=1: return a
    return ((1-inverse(m,a)*m)//a)%m
 
def lowbit(n):
    return n&-n
 
class BIT:
    def __init__(self,arr):
        self.arr=arr
        self.n=len(arr)-1
        
    def update(self,x,v):
        while x<=self.n:
            self.arr[x]+=v
            x+=x&-x
 
    def query(self,x):
        ans=0
        while x:
            ans+=self.arr[x]
            x&=x-1
        return ans

class ST:
    def __init__(self,arr):#n!=0
        n=len(arr)
        mx=n.bit_length()#取不到
        self.st=[[0]*mx for i in range(n)]
        for i in range(n):
            self.st[i][0]=arr[i]
        for j in range(1,mx):
            for i in range(n-(1<<j)+1):
                self.st[i][j]=max(self.st[i][j-1],self.st[i+(1<<j-1)][j-1])
    def query(self,l,r):
        if l>r:return -inf
        s=(r+1-l).bit_length()-1
        return max(self.st[l][s],self.st[r-(1<<s)+1][s])
        
class DSU:#容量+路径压缩
    def __init__(self,n):
        self.c=[-1]*n
 
    def same(self,x,y):
        return self.find(x)==self.find(y)
 
    def find(self,x):
        if self.c[x]<0:
            return x
        self.c[x]=self.find(self.c[x])
        return self.c[x]
 
    def union(self,u,v):
        u,v=self.find(u),self.find(v)
        if u==v:
            return False
        if self.c[u]>self.c[v]:
            u,v=v,u
        self.c[u]+=self.c[v]
        self.c[v]=u
        return True
 
    def size(self,x): return -self.c[self.find(x)]
    
class UFS:#秩+路径
    def __init__(self,n):
        self.parent=[i for i in range(n)]
        self.ranks=[0]*n
 
    def find(self,x):
        if x!=self.parent[x]:
            self.parent[x]=self.find(self.parent[x])
        return self.parent[x]
 
    def union(self,u,v):
        pu,pv=self.find(u),self.find(v)
        if pu==pv:
            return False
        if self.ranks[pu]>=self.ranks[pv]:
            self.parent[pv]=pu
            if self.ranks[pv]==self.ranks[pu]:
                self.ranks[pu]+=1
        else:
            self.parent[pu]=pv

class UF:#秩+路径+容量，边数
    def __init__(self,n):
        self.parent=[i for i in range(n)]
        self.ranks=[0]*n
        self.size=AI(n,1)
        self.edge=A(n)
 
    def find(self,x):
        if x!=self.parent[x]:
            self.parent[x]=self.find(self.parent[x])
        return self.parent[x]
 
    def union(self,u,v):
        pu,pv=self.find(u),self.find(v)
        if pu==pv:
            self.edge[pu]+=1
            return False
        if self.ranks[pu]>=self.ranks[pv]:
            self.parent[pv]=pu
            self.edge[pu]+=self.edge[pv]+1
            self.size[pu]+=self.size[pv]
            if self.ranks[pv]==self.ranks[pu]:
                self.ranks[pu]+=1
        else:
            self.parent[pu]=pv
            self.edge[pv]+=self.edge[pu]+1
            self.size[pv]+=self.size[pu]
 
def Prime(n):
    c=0
    prime=[]
    flag=[0]*(n+1) 
    for i in range(2,n+1):
        if not flag[i]:
            prime.append(i)
            c+=1
        for j in range(c):
            if i*prime[j]>n: break
            flag[i*prime[j]]=prime[j]
            if i%prime[j]==0: break
    return flag
 
def dij(s,graph):
    d=AI(n,inf)
    d[s]=0
    heap=[(0,s)]
    vis=A(n)
    while heap:
        dis,u=heappop(heap)
        if vis[u]:
            continue
        vis[u]=1
        for v,w in graph[u]:
            if d[v]>d[u]+w:
                d[v]=d[u]+w
                heappush(heap,(d[v],v))
    return d

def bell(s,g):#bellman-Ford
    dis=AI(n,inf)
    dis[s]=0
    for i in range(n-1):
        for u,v,w in edge:
            if dis[v]>dis[u]+w:
                dis[v]=dis[u]+w
    change=A(n)
    for i in range(n):
        for u,v,w in edge:
            if dis[v]>dis[u]+w:
                dis[v]=dis[u]+w
                change[v]=1
    return dis

def lcm(a,b): return a*b//gcd(a,b)
def lis(nums):
    res=[]
    for k in nums:
        i=bisect.bisect_left(res,k)
        if i==len(res):
            res.append(k)
        else:
            res[i]=k
    return len(res)

def RP(nums):#逆序对
    n = len(nums)
    s=set(nums)
    d={}
    for i,k in enumerate(sorted(s),1):
        d[k]=i
    bi=BIT([0]*(len(s)+1))
    ans=0
    for i in range(n-1,-1,-1):
        ans+=bi.query(d[nums[i]]-1)
        bi.update(d[nums[i]],1)
    return ans
 
class DLN:
    def __init__(self,val):
        self.val=val
        self.pre=None
        self.next=None

def nb(i,j,n,m):
    for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
        if 0<=ni<n and 0<=nj<m:
            yield ni,nj

def topo(n):
    q=deque()
    res=[]
    for i in range(1,n+1):
        if ind[i]==0:
            q.append(i)
            res.append(i)
    while q:
        u=q.popleft()
        for v in g[u]:
            ind[v]-=1
            if ind[v]==0:
                q.append(v)
                res.append(v)
    return res

@bootstrap
def gdfs(r,p):
    for ch in g[r]:
        if ch!=p:
            yield gdfs(ch,r)
    yield None

@bootstrap
def dfs(r,p):
    vis[r]=1
    par[r]=p
    #print(r,p,par)
    for ch in g[r]:
        if not vis[ch]:
            #print(r,ch)
            g2[r]+=1
            yield dfs(ch,r)
    yield None
        
#from random import randint


t=1
for i in range(t):
    n,m,X=RL()
    a=RLL()
    g=G(n)
    #pq=a.copy()
    #heapify(pq)
    edge={}
    g=G(n)
    g2=A(n)
    par=AI(n,-1)
    for i in range(m):
        x,y=RL()
        x-=1
        y-=1
        if x>y:x,y=y,x
        edge[x,y]=i+1
        g[x].append(y)
        g[y].append(x)
    if sum(a)<(n-1)*X:
        print('NO')
        exit()
    vis=A(n)
    dfs(0,-1)
    #print(par)
    q=deque()
    ans=[]
    back=[]
    for i in range(n):
        if not g2[i]:
            q.append(i)
    #print(q,par,g2)
    while q:
        u=q.popleft()
        v=par[u]
        if u>v:
            x,y=v,u
        else:
            x,y=u,v
        if v==-1:
            break
        if a[u]+a[v]>=X:
            a[v]=a[u]+a[v]-X
            ans.append(edge[x,y])
        else:
            back.append(edge[x,y])
        g2[v]-=1
        if g2[v]==0:
            q.append(v)
    back.reverse()
    ans=ans+back
    print('YES')
    for ind in ans:
        print(ind)
            
    #print(ans)
   

        



                 
    
    
''' 
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()
'''