Chef has two integer sequences $A_1, A_2, \ldots, A_N$ and $B_1, B_2, \ldots, B_M$. You should choose $N+M-1$ pairs, each in the form $(A_x, B_y)$, such that the sums $A_x + B_y$ are all pairwise distinct.
It is guaranteed that under the given constraints, a solution always exists. If there are multiple solutions, you may find any one.
-----Input-----
- The first line of the input contains two space-separated integers $N$ and $M$.
- The second line contains $N$ space-separated integers $A_1, A_2, \ldots, A_N$.
- The third line contains $M$ space-separated integers $B_1, B_2, \ldots, B_M$.
-----Output-----
Print $N+M-1$ lines. Each of these lines should contain two space-separated integers $x$ and $y$ denoting that you chose a pair $(A_{x+1}, B_{y+1})$.
-----Constraints-----
- $1 \le N, M \le 2 \cdot 10^5$
- $|A_i| \le 10^9$ for each valid $i$
- $|B_i| \le 10^9$ for each valid $i$
- $A_1, A_2, \ldots, A_N$ are pairwise distinct
- $B_1, B_2, \ldots, B_M$ are pairwise distinct
-----Subtasks-----
Subtask #1 (25 points): $1 \le N, M \le 10^3$
Subtask #2 (75 points): original constraints
-----Example Input-----
3 2
10 1 100
4 3
-----Example Output-----
2 1
0 0
1 0
0 1
-----Explanation-----
The chosen pairs and their sums are:
- $A_3 + B_2 = 100+3 = 103$
- $A_1 + B_1 = 10+4 = 14$
- $A_2 + B_1 = 1+4 = 5$
- $A_1 + B_2 = 10+3 = 13$
Since all sums are distinct, this is a correct output.
N,M = [int(i) for i in input().split()]
A = [int(i) for i in input().split()]
B = [int(i) for i in input().split()]

ans = []

seen = {}
flag = False

for i in range(N):
    for j in range(M):
        if(len(ans)==N+M-1):
            flag = True
            break
        x = A[i]+B[j]
        if(x not in seen):
            ans.append('{} {}'.format(i,j))
            seen[x] = True
    if(flag):
        break

for i in ans:
    print(i)
N,M = [int(i) for i in input().split()]
A = [int(i) for i in input().split()]
B = [int(i) for i in input().split()]

ans = []

seen = {}
flag = False

for i in range(N):
    for j in range(M):
        if(len(ans)==N+M-1):
            flag = True
            break
        x = A[i]+B[j]
        if(x not in seen):
            ans.append('{} {}'.format(i,j))
            seen[x] = True
    if(flag):
        break

for i in ans:
    print(i)