Evlampiy was gifted a rooted tree. The vertices of the tree are numbered from 1 to n. Each of its vertices also has an integer a_i written on it. For each vertex i, Evlampiy calculated c_i — the number of vertices j in the subtree of vertex i, such that a_j < a_i. 

<image>Illustration for the second example, the first integer is a_i and the integer in parentheses is c_i

After the new year, Evlampiy could not remember what his gift was! He remembers the tree and the values of c_i, but he completely forgot which integers a_i were written on the vertices.

Help him to restore initial integers!

Input

The first line contains an integer n (1 ≤ n ≤ 2000) — the number of vertices in the tree.

The next n lines contain descriptions of vertices: the i-th line contains two integers p_i and c_i (0 ≤ p_i ≤ n; 0 ≤ c_i ≤ n-1), where p_i is the parent of vertex i or 0 if vertex i is root, and c_i is the number of vertices j in the subtree of vertex i, such that a_j < a_i.

It is guaranteed that the values of p_i describe a rooted tree with n vertices.

Output

If a solution exists, in the first line print "YES", and in the second line output n integers a_i (1 ≤ a_i ≤ {10}^{9}). If there are several solutions, output any of them. One can prove that if there is a solution, then there is also a solution in which all a_i are between 1 and 10^9.

If there are no solutions, print "NO".

Examples

Input


3
2 0
0 2
2 0


Output


YES
1 2 1 

Input


5
0 1
1 3
2 1
3 0
2 0


Output


YES
2 3 2 1 2
def ddfs(root):
    global count
    global flag
    visit=[0]*(n+1)
    stack=[0,root]
    stack2=[]
    while(len(stack)>1):
        par=stack.pop()
        stack2.append(par)
        if(visit[par]):
            continue
        visit[par]=1
        for i in graph[par]:
            if(visit[i]==0):
                stack.append(i)
    visited=[0]*(n+1)
    while(len(stack2)>1):
        q = stack2.pop()
        if(count[q]<under[q]):
            flag=1
        visited[q]=1
        for i in graph[q]:
            if(visited[i]==0):
                count[i]+=count[q]+1


def dfs(root):
    stack=[root]
    visit=[0]*(n+1)

    while(stack):
        par=stack.pop()
        if(visit[par]):
            continue
        yy = seg_queries(under[par]+1,store - 1)
        final[par]=yy
        seg_update(store-1+yy)
        visit[par]=1
        for i in graph[par]:
            if(visit[i]==0):
                stack.append(i)

def seg_update(yy):
    seg_tree[yy]=0
    while(yy>1):
        yy=(yy>>1)
        seg_tree[yy]=seg_tree[yy<<1]+seg_tree[(yy<<1)+1]

def seg_queries(p,y):
    x=1
    if(seg_tree[x]<p):
        return -1
    while(x<=y):
        if(seg_tree[x<<1]>=p):
            x=x<<1
        else:
            p -= seg_tree[x << 1]
            x=(x<<1)+1

    return (x-y)

def seg_build(x):
    while(x>=1):
        seg_tree[x]=seg_tree[x<<1]+seg_tree[(x<<1)+1]
        x-=1

n=int(input())
flag=0
graph=[[] for i in range(n+1)]
under=[0 for i in range(n+1)]
for i in range(1,n+1):
    a,b=map(int,input().split())
    under[i]=b
    if(a==0):
        root=i
    else:
        graph[a].append(i)
        graph[i].append(a)
count=[0]*(n+1)
ddfs(root)
i=1
while(i<n):
    i=i<<1
store=i
seg_tree=[0 for j in range(store<<1)]
for j in range(store,store+n):
    seg_tree[j]=1
final=[0 for i in range(n+1)]
seg_build(store-1)
dfs(root)
if(flag==0):
    print("YES")
    print(*final[1:])
else:
    print("NO")