Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.

Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input

The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.

Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).

Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).

The numbers in the lines are separated by single spaces.

Output

On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Examples

Input

3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3


Output

9 18 17


Input

1 1 1
1
1 1 1
1 1


Output

2


Input

4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3


Output

5 18 31 20
maxn = 1e5
n,m,k=input().split(' '); n=int(n); m=int(m); k=int(k)# 输入第一行
a = input().split(' ')# 输入第二行
for i in range(len(a)): a[i]=int(a[i])
for i in range(len(a)-1,0,-1): a[i]-=a[i-1] # 差分
a.append(0)
L = []; R = []; d = []; cnt=[]
for i in range(m):# 读入所有操作
    t = input().split(' ')
    L.append(int(t[0])); 
    R.append(int(t[1])); 
    d.append(int(t[2])); 
    cnt.append(0)
cnt.append(0)
for i in range(k):# 读入所有查询
    x,y=input().split(' '); x=int(x); y=int(y)
    cnt[x-1]+=1; cnt[y]-=1
for i in range(1,m):# 前缀和
    cnt[i]+=cnt[i-1]# 每个操作被执行的次数
for i in range(m): # 遍历操作，每个操作执行应有的次数
    a[L[i]-1]+=d[i]*cnt[i]
    a[R[i]]-=d[i]*cnt[i]
for i in range(1,n):
    a[i]+=a[i-1]
for i in range(n):
    print(a[i],end=" ")
print("")
