#P1411E. Poman Numbers

    ID: 2955 Type: RemoteJudge 1000ms 256MiB Tried: 0 Accepted: 0 Difficulty: 8 Uploaded By: Tags>bitmasksgreedymathstrings*2300

Poman Numbers

Description

You've got a string SS consisting of nn lowercase English letters from your friend. It turned out that this is a number written in poman numerals. The poman numeral system is long forgotten. All that's left is the algorithm to transform number from poman numerals to the numeral system familiar to us. Characters of SS are numbered from 11 to nn from left to right. Let's denote the value of SS as f(S)f(S), it is defined as follows:

  • If S>1|S| > 1, an arbitrary integer mm (1m<S1 \le m < |S|) is chosen, and it is defined that f(S)=f(S[1,m])+f(S[m+1,S])f(S) = -f(S[1, m]) + f(S[m + 1, |S|]), where S[l,r]S[l, r] denotes the substring of SS from the ll-th to the rr-th position, inclusively.
  • Otherwise S=cS = c, where cc is some English letter. Then f(S)=2pos(c)f(S) = 2^{pos(c)}, where pos(c)pos(c) is the position of letter cc in the alphabet (pos(pos(a)=0) = 0, pos(pos(z)=25) = 25).

Note that mm is chosen independently on each step.

Your friend thinks it is possible to get f(S)=Tf(S) = T by choosing the right mm on every step. Is he right?

The first line contains two integers nn and TT (2n1052 \leq n \leq 10^5, 1015T1015-10^{15} \leq T \leq 10^{15}).

The second line contains a string SS consisting of nn lowercase English letters.

Print "Yes" if it is possible to get the desired value. Otherwise, print "No".

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

Input

The first line contains two integers nn and TT (2n1052 \leq n \leq 10^5, 1015T1015-10^{15} \leq T \leq 10^{15}).

The second line contains a string SS consisting of nn lowercase English letters.

Output

Print "Yes" if it is possible to get the desired value. Otherwise, print "No".

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

Sample Input 1

2 -1
ba

Sample Output 1

Yes

Sample Input 2

3 -7
abc

Sample Output 2

No

Sample Input 3

7 -475391
qohshra

Sample Output 3

Yes

Note

In the second example, you cannot get 7-7. But you can get 11, for example, as follows:

  1. First choose m=1m = 1, then f(f(abc)=f() = -f(a)+f() + f(bc))
  2. f(f(a)=20=1) = 2^0 = 1
  3. f(f(bc)=f() = -f(b)+f() + f(c)=21+22=2) = -2^1 + 2^2 = 2
  4. In the end f(f(abc)=1+2=1) = -1 + 2 = 1