Codeforces Educational Codeforces Round 31

B. Japanese Crosswords Strike Back

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A one-dimensional Japanese crossword can be represented as a binary string of length x. An encoding of this crossword is an array a of size n, where n is the number of segments formed completely of 1's, and ai is the length of i-th segment. No two segments touch or intersect.

For example: 

• If x = 6 and the crossword is 111011, then its encoding is an array {3, 2}; 
• If x = 8 and the crossword is 01101010, then its encoding is an array {2, 1, 1}; 
• If x = 5 and the crossword is 11111, then its encoding is an array {5}; 
• If x = 5 and the crossword is 00000, then its encoding is an empty array. 

Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!

Input

The first line contains two integer numbers n and x (1 ≤ n ≤ 100000, 1 ≤ x ≤ 109) — the number of elements in the encoding and the length of the crossword Mishka picked.

The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 10000) — the encoding.

Output

Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.

Examples

input

2 41 3

output

NO

input

3 103 3 2

output

YES

input

2 101 3

output

NO

题意：有一个长度为x的01串，其中含有n个连续1子串，每个连续1子串中1的个数为ai，问是否只存在一种情况使其成立

思路：只有一种情况时就是在相邻两个连续1子串中只能含有一个0，只有一个连续1子串则整个串中不能含有0

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;#define LL long longint main(){    LL n,x,a,sum = 0;    scanf("%lld%lld",&n,&x);    for(int i=1;i<=n;i++){        scanf("%lld",&a);        sum+=a;        if(i!=n) sum++;    }    if(sum==x) printf("YES\n");    else printf("NO\n");    return 0;}