Codeforces Round #635 B. Kana and Dragon Quest game（贪心）

Kana was just an ordinary high school girl before a talent scout discovered her. Then, she became an idol. But different from the stereotype, she is also a gameholic.

One day Kana gets interested in a new adventure game called Dragon Quest. In this game, her quest is to beat a dragon.

 

The dragon has a hit point of xx initially. When its hit point goes to 00 or under 00 , it will be defeated. In order to defeat the dragon, Kana can cast the two following types of spells.

• Void Absorption

  Assume that the dragon's current hit point is hh , after casting this spell its hit point will become ⌊h2⌋+10⌊h2⌋+10 . Here ⌊h2⌋⌊h2⌋ denotes hh divided by two, rounded down.

• Lightning Strike

  This spell will decrease the dragon's hit point by 1010 . Assume that the dragon's current hit point is hh , after casting this spell its hit point will be lowered to h−10h−10 .

Due to some reasons Kana can only cast no more than nn Void Absorptions and mm Lightning Strikes. She can cast the spells in any order and doesn't have to cast all the spells. Kana isn't good at math, so you are going to help her to find out whether it is possible to defeat the dragon.

Input

The first line contains a single integer tt (1≤t≤10001≤t≤1000 )  — the number of test cases.

The next tt lines describe test cases. For each test case the only line contains three integers xx , nn , mm (1≤x≤1051≤x≤105 , 0≤n,m≤300≤n,m≤30 )  — the dragon's intitial hit point, the maximum number of Void Absorptions and Lightning Strikes Kana can cast respectively.

Output

If it is possible to defeat the dragon, print "YES" (without quotes). Otherwise, print "NO" (without quotes).

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

Example

Input

Copy

7
100 3 4
189 3 4
64 2 3
63 2 3
30 27 7
10 9 1
69117 21 2

Output

Copy

YES
NO
NO
YES
YES
YES
YES
优先使用第一种攻击，直到血量不降反升之前，再用第二种攻击，看看最后可不可以KO。

#include <bits/stdc++.h>
using namespace std;
int x,n,m;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        scanf("%d%d%d",&x,&n,&m);
        if(x<=m*10)
        {
            cout<<"YES"<<endl;
            continue;
        }
        while(n>0)
        {
            if(x/2+10<x)
            {
                x=x/2+10;
                n--;
            }
            else break;
        }
        if(x<=m*10)
        {
            cout<<"YES"<<endl;
        }
        else cout<<"NO"<<endl;
    }
    return 0;
}