D. Find the Different Ones!

time limit per test

5 seconds

memory limit per test

256 megabytes

You are given an array a of n integers, and q queries.

Each query is represented by two integers l and r (1≤l≤r≤n). Your task is to find, for each query, two indices i and j (or determine that they do not exist) such that:

l≤i≤r;
l≤j≤r;
ai≠aj.

In other words, for each query, you need to find a pair of different elements among al,al+1,...,ar, or report that such a pair does not exist.

Input

The first line of the input contains a single integer t (1≤t≤104) --- the number of test cases. The descriptions of the test cases follow.

The first line of each test case contains a single integer n (2≤n≤2⋅105) --- the length of the array a.

The second line of each test case contains n integers a1,a2,...,an (1≤ai≤106) --- the elements of the array a.

The third line of each test case contains a single integer q (1≤q≤2⋅105) --- the number of queries.

The next q lines contain two integers each, l and r (1≤l<r≤n) --- the boundaries of the query.

It is guaranteed that the sum of the values of n across all test cases does not exceed 2⋅105. Similarly, it is guaranteed that the sum of the values of q across all test cases does not exceed 2⋅105.

Output

For each query, output two integers separated by space: i and j (l≤i,j≤r), for which ai≠aj. If such a pair does not exist, output i=−1 and j=−1.

You may separate the outputs for the test cases with empty lines. This is not a mandatory requirement.

Example

Input

Copy

复制代码

1 1 2 1 1

1 5

1 2

1 3

30 20 20 10 10 20

1 2

2 3

2 4

2 6

3 5

5 2 3 4

1 2

1 4

2 3

2 4

1 4 3 2 4

1 5

2 4

3 4

3 5

4 5

2 3 1 4 2

1 2

1 4

1 5

2 4

2 5

3 5

4 5

Output

Copy

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解题说明：此题是一道模拟题，可以采用动态规划算法，假设next_diff[i] 等于 i的左边第一个与arr[i]不一样的点的下标。如果 arr[i]!=arr[i-1] , next_diff[i]=i-1; 否则 next_diff[i]=next_diff[i-1];

如果，第i个数与第i-1个数不一样，记录这个不一样的地方的开端：i-1

如果i+1还是不一样的话，那么记录不一样的开端：i

如果后面都一样，那么后面都等于 i

如果区间[l,r] 之间所有数都一样，那么 next_diff[l]=next_diff[r]

否则，我们只要输出 r 和next_diff[r]即可

cpp 复制代码

#include <stdio.h>

int main()
{
	int t;
	scanf("%d", &t);
	while (t--) 
	{
		int n, q;
		scanf("%d", &n);
		int arr[200005], next_diff[200005];
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &arr[i]);
		}
		next_diff[n - 1] = n;
		for (int i = n - 2; i >= 0; i--) 
		{
			if (arr[i] == arr[i + 1])
			{
				next_diff[i] = next_diff[i + 1];
			}
			else
			{
				next_diff[i] = i + 1;
			}
		}

		scanf("%d", &q);
		while (q--)
		{
			int l, r;
			scanf("%d %d", &l, &r);
			l--; 
			r--;
			if (next_diff[l] > r)
			{
				printf("-1 -1\n");
			}
			else
			{
				printf("%d %d\n", l + 1, next_diff[l] + 1);
			}
		}
	}
	return 0;
}