There are ~n~ people in a class, numbered ~1~ through ~n~. Friendship is mutual: if ~a~ is friends with ~b~, then ~b~ is friends with ~a~. Two people are in the same friend group if they are the same person, or if they are connected by a chain of friendships (friends, friends of friends, and so on).

At the start of the year, some friendships exist. Over the year, those friendships break one by one until none remain. You are given a chronological log of ~q~ events. Every friendship that ever existed appears in this log as a breakup, so the initial friendships are exactly the pairs that later break.

Process the events in order and answer queries about who is still in the same friend group.

Input

The first line contains two integers ~n~ and ~q~, the number of people and the number of events.

The next ~q~ lines each describe one event in one of the following forms:

• 1 a b — the friendship between ~a~ and ~b~ breaks (and is removed). If that friendship is already gone, nothing happens.
• 2 a b — ask whether ~a~ and ~b~ are currently in the same friend group.

Output

For each event of type ~2~, output one line:

• Yes if ~a~ and ~b~ are in the same friend group,
• No otherwise.

Constraints

• ~1 \le n \le 10^5~
• ~1 \le q \le 5 \times 10^5~
• ~1 \le a, b \le n~
• ~a \ne b~

Example 1

Input

5 10
1 1 2
1 4 5
2 3 4
1 3 4
1 2 5
2 2 5
2 2 5
1 3 5
2 2 4
1 1 5

Output

Yes
No
No
No

Explanation

The breakups in the log are the pairs ~(1, 2)~, ~(4, 5)~, ~(3, 4)~, ~(2, 5)~, ~(3, 5)~, and ~(1, 5)~, so those six friendships exist at the start.

• After ~(1, 2)~ and ~(4, 5)~ break, ~3~ and ~4~ are still friends directly, so the answer is Yes.
• After ~(3, 4)~ and ~(2, 5)~ also break, ~2~ and ~5~ are no longer connected, so both queries answer No.
• After ~(3, 5)~ breaks, ~2~ and ~4~ are still not connected, so the answer is No.

Example 2

Input

10 20
2 3 9
2 6 10
1 3 10
2 4 8
1 1 9
1 2 8
2 6 10
2 4 8
1 2 9
1 6 10
1 5 8
1 2 8
1 5 6
2 7 9
1 8 10
2 7 9
2 2 4
1 3 5
1 5 8
2 1 8

Output

Yes
Yes
No
Yes
No
No
No
No
No