题解 | #KiKi判断上三角矩阵#

n=int(input())
list1=[]
for i in range(n):
    a=list(map(int,input().split()))
    list1.append(a)
num=0
for j in range(1,n):
    for k in range(j):
        if list1[j][k]==0:
            num+=1
        else:
            break
if num==n*(n-1)//2:
    print("YES")
else:
    print("NO")