Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.

Input

The first line contains an integer n (1 ≤ n ≤ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.

Output

If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 ≤ l < r ≤ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.

Examples

Input

8
1 6 5 4 3 2 7 8


Output

2 6


Input

4
2 3 4 1


Output

0 0


Input

4
1 2 3 4


Output

0 0
x=int(raw_input())
d=raw_input()
d=d.split()
res=''
xc=0
xb=0
k=0
v=0
j=0
for i in range(0,x):
    d[i]=int(d[i])
l=sorted(d)
for i in range(0,len(d)):
    if(d[i]!=l[i]):
        k=i
        j=j+1
        break
for i in range(len(d)-1,-1,-1):
    if(d[i]!=l[i]):
        v=i
        j=j+1
        break
for i in range(k,v+1):
    xc=xc+1
r=[0]*xc
so=[0]*xc
for i in range(k,v+1):
    
    so[xb]=l[i]
    xb=xb+1
xb=0    
for i in range(v,k-1,-1):
    r[xb]=d[i]
    xb=xb+1


if(r==so)&(j>0):
    res=str(l[k])+" "+str(l[v])
    print(res)
else:
    print("0"+" "+"0")
          
    
