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 4 1 3
Output
NO
Input
3 10 3 3 2
Output
YES
Input
2 10 1 3
Output
NO
题意：输入n，m表示有一个长为m的01串，判断是否可以按要求组成合法的01串，n表示01串中 1区间的个数，接着输入n个数，表示这些1区间的长度。0只能出先在两个1区间之间！。
思路：0只能出现在1区间之间，也就是说，两个0不能相邻，那么只有当0的个数刚好等于1区间的个数减一才满足条件。
#include "iostream"
using namespace std;
int main()
{
ios::sync_with_stdio(false);
int n,x,t,sum=0;
cin>>n>>x;
for(int i=0;i<n;i++) {cin>>t;sum+=t;}
if(x-sum==n-1) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
return 0;
}