Problem: 2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.

For example, this picture describes the chronological order of games with k = 3:

<image>

Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:

  * if s_i is 0, then the team with lower index wins the i-th game; 
  * if s_i is 1, then the team with greater index wins the i-th game; 
  * if s_i is ?, then the result of the i-th game is unknown (any team could win this game). 



Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.

You are given the initial state of the string s. You have to process q queries of the following form: 

  * p c — replace s_p with character c, and print f(s) as the result of the query. 

Input

The first line contains one integer k (1 ≤ k ≤ 18).

The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.

The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.

Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.

Output

For each query, print one integer — f(s).

Example

Input


3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1


Output


1
2
3
3
5
4

Solution (in CPP):

Solution:
//JAI MATA DI _/\_
//created by janwar!
//You say debug I will hear trouble.
//Sometimes code works acidentally so don't ask me the explantation of it.
 
#include<bits/stdc++.h>
 
using namespace std;
 
#define                                   ll long long 
#define                                   ld long double
#define                                   ull unsigned long long
#define                                   scn(n) scanf("%d",&n)
#define                                   lscn(n) scanf("%lld",&n)
#define                                   lpri(n) printf("%lld",n)
#define                                   pri(n) printf("%d",n)
#define                                   pln() printf("\n")
#define                                   priln(n) printf("%d\n",n)
#define                                   lpriln(n) printf("%lld\n",n)
#define                                   rep(i,init,n) for(int i=init;i<n;i++)
#define                                   rev(i,n) for(int i=n-1;i>=0;i--)
#define                                   pb push_back     
#define                                   mp make_pair
#define                                   F first
#define                                   S second
#define                                   gcd __gcd
#define                                   inf INT_MAX
#define                                   ninf INT_MIN   
#define                                   inf INT_MAX
#define                                   lmax LLONG_MAX
#define                                   lmin -100000000000000ll 
#define                                   tc ll ttt; cin>>ttt; while(ttt--)
#define                                   VC vector<ll> 
#define                                   MP map<ll,ll> 
#define                                   lb lower_bound
#define                                   endl "\n"
#define                                   ub upper_bound
#define                                   all(a) a.begin(),a.end()
#define                                   MC map<char,ll>
#define                                   VP vector<pair<ll,ll>>
#define                                   read(a,n)  rep(i,0,n)cin>>a[i];
#define                                   sortt(v)   sort(v.begin(),v.end());
#define                                   fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define                                   set_bits __builtin_popcountll
 
// typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag, tree_order_statistics_node_update > pbds; // find_by_order, order_of_key
 
const int mod = 1e9+7;
const int N =  1e6 +4;
 
 
 
#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif
 
void _print(ll t) {cerr << t;}
void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(ld t) {cerr << t;}
void _print(double t) {cerr << t;}
void _print(ull t) {cerr << t;}
 
template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.F); cerr << ","; _print(p.S); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}








void JAI_SHREE_RAM()
{
ll k; cin >> k;



string s; cin >> s;

s+='_'; // making it one based.
reverse(all(s)); //reverse it to make it  seg tree.

ll N=1<<(k+1);
VC dp(N); 
ll n=1<<k;
for(int i=n;i<N;i++)
  dp[i]=1;





for(int i=n-1;i>0;i--)
{

   if(s[i]=='0')
    dp[i]=dp[2*i+1];  // ans after reversing ans lies in right child
  else
    if(s[i]=='1')
      dp[i]= dp[2*i];  // after reversing ans lies in left child.
    else
      dp[i]= dp[2*i+1]+ dp[2*i];  // taking account of both in case of ?
}
debug(dp);
ll q; cin >> q;

while(q--)
{
  ll indx; char c;
  cin >> indx; cin >> c;
  indx= (1<<(k))-indx;

  s[indx]=c;

  while(indx)
  {
    if(s[indx]=='0')
      dp[indx]= dp[2*indx+1];     // updating.
    else
      if(s[indx]=='1')
        dp[indx]=dp[2*indx];
      else
        dp[indx]=dp[2*indx]+ dp[2*indx+1];

      indx/=2;
  }
  debug(dp);
cout<<dp[1]<<endl;  // printing dp[1] number of possible ways.
}

}



     
  














 
int main() {
#ifndef ONLINE_JUDGE
  freopen("Error.txt", "w", stderr);
#endif
fast;
ll tttt=1;







// cin >> tttt;
while(tttt--)
{
 // solve();
 
  JAI_SHREE_RAM();
}
 
 
 
}