Given an array $a_1, a_2, \dots, a_n$, you need to handle a total of $q$ updates and queries of two types:

$1$ $l$ $r$ — for each index $i$ with $l \leq i \leq r$, update the value of $a_i$ to the sum of the digits of $a_i$.

$2$ $x$ — output $a_x$.


-----Input-----

The first line of the input contains an integer $t$ ($1 \leq t \leq 1000$) — the number of testcases.

The first line of each test case contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the size of the array and the number of queries, respectively.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$).

The next $q$ lines of each test case are of two forms:

$1$ $l$ $r$ ($1 \leq l \leq r \leq n$) — it means, for each index $i$ with $l \leq i \leq r$, you should update the value of $a_i$ to the sum of its digits.

$2$ $x$ ($1 \leq x \leq n$) — it means you should output $a_x$.

There is at least one query of the second type.

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

The sum of $q$ over all test cases does not exceed $2 \cdot 10^5$.


-----Output-----

For each test case, output the answers of queries of the second type, in the order they are given.


-----Examples-----

Input
3
5 8
1 420 69 1434 2023
1 2 3
2 2
2 3
2 4
1 2 5
2 1
2 3
2 5
2 3
9999 1000
1 1 2
2 1
2 2
1 1
1
2 1
Output
6
15
1434
1
6
7
36
1
1


-----Note-----

In the first test case, the following process occurs:

Initially, $a = [1, 420, 69, 1434, 2023]$.

The operation is performed for $l=2$, $r=3$, yielding $[1, {6}, {15}, 1434, 2023]$.

We are queried for $x=2$, $x=3$, and $x=4$, and output $6$, $15$, and $1434$.

The operation is performed for $l=2$, $r=5$, yielding $[1, {6}, {6}, {12}, {7}]$.

We are queried for $x=1$, $x=3$, and $x=5$, and output $1$, $6$, and $7$.

Difficulty: MEDIUM

Solutions:
import sys
input = sys.stdin.readline
output = sys.stdout.write

def find(x):
	t = x
	while fa[x] != x:
		x = fa[x]
	while t != x:
		(fa[t], t) = (x, fa[t])
	return x

def union(son, father):
	x = find(son)
	y = find(father)
	fa[x] = y

def f(num):
	ans = 0
	while num > 0:
		ans += num % 10
		num //= 10
	return ans
n = int(input())
for _ in range(n):
	(m, q) = list(map(int, input().split()))
	fa = list(range(m + 1))
	if fa[0] == 9999999:
		print(0)
		break
	a = list(map(int, input().split()))
	for _ in range(q):
		t = list(map(int, input().split()))
		if t[0] == 1:
			(l, r) = (t[1], t[2])
			l -= 1
			l = fa[l]
			while l < r:
				a[l] = f(a[l])
				if a[l] < 10:
					union(l, l + 1)
				l = fa[l + 1]
		else:
			print(a[t[1] - 1])