You are given a long decimal number a consisting of n digits from 1 to 9. You also have a function f that maps every digit from 1 to 9 to some (possibly the same) digit from 1 to 9.

You can perform the following operation no more than once: choose a non-empty contiguous subsegment of digits in a, and replace each digit x from this segment with f(x). For example, if a = 1337, f(1) = 1, f(3) = 5, f(7) = 3, and you choose the segment consisting of three rightmost digits, you get 1553 as the result.

What is the maximum possible number you can obtain applying this operation no more than once?

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of digits in a.

The second line contains a string of n characters, denoting the number a. Each character is a decimal digit from 1 to 9.

The third line contains exactly 9 integers f(1), f(2), ..., f(9) (1 ≤ f(i) ≤ 9).

Output

Print the maximum number you can get after applying the operation described in the statement no more than once.

Examples

Input


4
1337
1 2 5 4 6 6 3 1 9


Output


1557


Input


5
11111
9 8 7 6 5 4 3 2 1


Output


99999


Input


2
33
1 1 1 1 1 1 1 1 1


Output


33
from sys  import stdin,stdout
 
import bisect
 
import math
 
from collections import deque
 
mod=10**9 +7 
 
def st():
    return list(stdin.readline().strip())
 
def inp():
    return int(stdin.readline())
 
def li():
    return list(map(int,stdin.readline().split()))
 
def mp():
    return map(int,stdin.readline().split())
 
def pr(n):
    stdout.write(str(n)+"\n")
    
def DFS(dictionary,vertex,visited):
    visited[vertex]=True
    stack=[vertex]
    print(vertex)
    while stack:
        a=stack.pop()
        for i in dictionary[a]:
            if not visited[i]:
                print(i)
                visited[i]=True
                stack.append(i)
 
 
def BFS(dictionary, vertex,visited):
    visited[vertex]=True
    q=deque()
    q.append(vertex)
    while q:
        a=q.popleft()
        for i in dictionary[a]:
            if not visited[i]:
                visited[i]=True
                q.append(i)
                print(i)
 
                
 
 
def soe(limit):
    l=[1]*(limit+1)
    l[0]=0
    l[1]=0
    prime=[]
    for i in range(2,limit+1):
        if l[i]:
            for j in range(i*i,limit+1,i):
                l[j]=0
    
    for i in range(2,limit+1):
        if l[i]:
            prime.append(i)
    return prime
 
def segsoe(low,high):
    limit=int(high**0.5)+1
    prime=soe(limit)
    n=high-low+1
    l=[0]*(n+1)
    for i in range(len(prime)):
        lowlimit=(low//prime[i])*prime[i]
        if lowlimit<low:
            lowlimit+=prime[i]
        if lowlimit==prime[i]:
            lowlimit+=prime[i]
        for j in range(lowlimit,high+1,prime[i]):
            l[j-low]=1
    for i in range(low,high+1):
        if not l[i-low]:
            if i!=1:
                print(i)
                
def gcd(a,b):
    while b:
        a=a%b
        b,a=a,b
    return a
 
def power(a,n):
    r=1
    while n:
        if n&1:
            r=(r*a)
        a*=a
        n=n>>1
    return r
    
def solve():
    n=inp()
    s=list(input())
    l=li()
    l=[0]+l
    i=0
    while i<n and int(s[i])>=l[int(s[i])]:
        i+=1
    while i<n and int(s[i])<=l[int(s[i])]:
        s[i]=str(l[int(s[i])])
        i+=1
    print(''.join(s))
        
        
            

 
for _ in range(1):
    solve()
