Description

You are given an integer x. Can you make x by summing up some number of 11,111,1111,11111,…? (You can use any number among them any number of times).

For instance,

33=11+11+11
144=111+11+11+11

Input

The first line of input contains a single integer t (1≤t≤10000) — the number of testcases.

The first and only line of each testcase contains a single integer x (1≤x≤109) — the number you have to make.

Output

For each testcase, you should output a single string. If you can make x, output “YES” (without quotes). Otherwise, output “NO”.

You can print each letter of “YES” and “NO” in any case (upper or lower).

Example

input

3
33
144
69

output

YES
YES
NO

Note
Ways to make 33 and 144 were presented in the statement. It can be proved that we can’t present 69 this way.

其实这个题目并不是很难，只需要稍微推导一下就好了

Codeforces Round #723 (Div. 2)B. I Hate 1111_i++


可以看到左右的数(1111…这种)都可以由111和11构成

所以我们只需要枚举111的数量，并且查看剩下的数能不能由11构成就好啦

ll ans;
int main() {
  int T = read;
  while(T --) {
    ll x; cin >> x;
    int flag = 0;
    for(itn i=0;i*111<=x;i++){
      if((x - i*111) % 11 == 0) {
        flag = 1;
        break;
      }
    }
    if(flag) puts("yes");
    else puts("no");
  }
  return 0;
}