Problem: We have a tree with N vertices. The vertices are numbered 1, 2, ..., N. The i-th (1 ≦ i ≦ N - 1) edge connects the two vertices A_i and B_i.

Takahashi wrote integers into K of the vertices. Specifically, for each 1 ≦ j ≦ K, he wrote the integer P_j into vertex V_j. The remaining vertices are left empty. After that, he got tired and fell asleep.

Then, Aoki appeared. He is trying to surprise Takahashi by writing integers into all empty vertices so that the following condition is satisfied:

* Condition: For any two vertices directly connected by an edge, the integers written into these vertices differ by exactly 1.



Determine if it is possible to write integers into all empty vertices so that the condition is satisfied. If the answer is positive, find one specific way to satisfy the condition.

Constraints

* 1 ≦ N ≦ 10^5
* 1 ≦ K ≦ N
* 1 ≦ A_i, B_i ≦ N (1 ≦ i ≦ N - 1)
* 1 ≦ V_j ≦ N (1 ≦ j ≦ K) (21:18, a mistake in this constraint was corrected)
* 0 ≦ P_j ≦ 10^5 (1 ≦ j ≦ K)
* The given graph is a tree.
* All v_j are distinct.

Input

The input is given from Standard Input in the following format:


N
A_1 B_1
A_2 B_2
:
A_{N-1} B_{N-1}
K
V_1 P_1
V_2 P_2
:
V_K P_K


Output

If it is possible to write integers into all empty vertices so that the condition is satisfied, print `Yes`. Otherwise, print `No`.

If it is possible to satisfy the condition, print N lines in addition. The v-th (1 ≦ v ≦ N) of these N lines should contain the integer that should be written into vertex v. If there are multiple ways to satisfy the condition, any of those is accepted.

Examples

Input

5
1 2
3 1
4 3
3 5
2
2 6
5 7


Output

Yes
5
6
6
5
7


Input

5
1 2
3 1
4 3
3 5
3
2 6
4 3
5 7


Output

No


Input

4
1 2
2 3
3 4
1
1 0


Output

Yes
0
-1
-2
-3

Solution (in PYTHON3):

Solution:

"""
Writer: SPD_9X2
https://atcoder.jp/contests/arc063/tasks/arc063_c

数字が置かれている頂点は範囲指定の制約になる
→制約をすべての点に伝播させれば、後はその範囲内から適当に選んで実装すればよい

一度根方向に制約を集め、それを全方位に伝播させればよい？
→それっぽさそう

偶奇に関してはちゃんと考えてね
→条件の両端の偶奇で判断可能
"""

import sys
sys.setrecursionlimit(5*10**5)

N = int(input())

lis = [ [] for i in range(N) ]

for i in range(N-1):

    a,b = map(int,input().split())
    a -= 1
    b -= 1

    lis[a].append(b)
    lis[b].append(a)


l = [float("-inf")] * N
r = [float("inf")]  * N

K = int(input())

for i in range(K):

    V,P = map(int,input().split())
    V -= 1
    l[V] = P
    r[V] = P

#dfsで条件を根方向に集める
def dfs(v,p):

    for nex in lis[v]:

        if nex != p:
            nl,nr = dfs(nex,v)

            if l[v] != float("-inf") and nl != float("-inf") and l[v] % 2 != nl % 2:
                print ("No")
                sys.exit()

            l[v] = max(l[v] , nl)
            r[v] = min(r[v] , nr)

            if l[v] > r[v]:
                print ("No")
                sys.exit()

    return l[v] - 1 , r[v] + 1

dfs(0,0)

#bfsで条件を配る
from collections import deque
q = deque([0])
unvisit = [True] * N
ans = [None] * N
ans[0] = l[0]

while len(q) > 0:
    
    v = q.popleft()
    unvisit[v] = False

    if l[v] > r[v]:
        print ("No")
        sys.exit()
    
    for nex in lis[v]:

        if unvisit[nex]:

            if l[nex] != float("-inf") and l[v] != float("-inf") and l[v] % 2 == l[nex] % 2:
                print ("No")
                sys.exit()

            l[nex] = max(l[nex] , l[v]-1)
            r[nex] = min(r[nex] , r[v]+1)
            ans[nex] = l[nex]

            q.append(nex)

print ("Yes")
for i in ans:
    print (i)