Interesting Subarray

For an array $a$ of integers let’s denote its maximal element as $\max(a)$, and minimal as $\min(a)$. We will call an array $a$ of $k$ integers interesting if $\max(a) – \min(a) \ge k$. For example, array $[1, 3, 4, 3]$ isn’t interesting as $\max(a) – \min(a) = 4 – 1 = 3 < 4$ while array $[7, 3, 0, 4, 3]$ is as $\max(a) – \min(a) = 7 – 0 = 7 \ge 5$.

You are given an array $a$ of $n$ integers. Find some interesting nonempty subarray of $a$, or tell that it doesn’t exist.

An array $b$ is a subarray of an array $a$ if $b$ can be obtained from $a$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end. In particular, an array is a subarray of itself.Input

The first line contains integer number $t$ ($1 \le t \le 10\,000$). Then $t$ test cases follow.

The first line of each test case contains a single integer $n$ ($2\le n \le 2\cdot 10^5$) — the length of the array.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($0\le a_i \le 10^9$) — the elements of the array.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.Output

For each test case, output “NO” in a separate line if there is no interesting nonempty subarray in $a$.

Otherwise, output “YES” in a separate line. In the next line, output two integers $l$ and $r$ ($1\le l \le r \le n$) — bounds of the chosen subarray. If there are multiple answers, print any.

You can print each letter in any case (upper or lower).Exampleinput

3
5
1 2 3 4 5
4
2 0 1 9
2
2019 2020

output

NO
YES
1 4
NO

Note

In the second test case of the example, one of the interesting subarrays is $a = [2, 0, 1, 9]$: $\max(a) – \min(a) = 9 – 0 = 9 \ge 4$.

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
      cin >> a[i];
}
bool found = false;
for (int i = 0; i < n - 1; i++) {
      if (abs(a[i] - a[i + 1]) > 1) {
cout << "YES" << '\n';
        cout << i + 1 << " " << i + 2 << '\n';
        found = true;
        break;
      }
    }
    if (!found) {
      cout << "NO" << '\n';
    }
  }
  return 0;
}