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:
#include<bits/stdc++.h>
using namespace std;
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
 
template<typename T,typename T1>T amax(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T amin(T &a,T1 b){if(b<a)a=b;return a;}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
#define ll long long
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define endl "\n"
#define ss second
#define pb push_back
#define pf push_front
#define pu push
#define pp pop_back
#define in insert
#define ff first
#define ss second
#define as assign
#define rs resize
#define rep(i,a,b) for(ll i = (a); i <(b); i++)
#define repp(i,a,b) for(ll i = (a); i > (b); i--)
#define all(x)  (x).begin(),(x).end()
#define rev(x) (x).rbegin(),(x).rend()
#define alll(it,a) for(auto it=(a).begin();it!=(a).end();it++)
#define revv(it,a) for(auto it=(a).rbegin();it!=(a).rend();it++)
#define ppc  __builtin_popcount
#define ppcll __builtin_popcountll
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define umap unordered_map
const ll mod = 1e9+7;
const long long INF=1e18;
/*sort(vec.begin(), vec.end(),
[](node const & e,  node const & g) -> bool{return e.mode < g.mode; } );
struct myComp{bool operator()(pll const& p1, pll const& p2)return p1.ff < p2.ff;}
priority_queue<pll, vector<pll>, greater<pll> > pq;
*max_element(v.begin(),v.begin()+j+1);*/
// const ll MOD = 998244353; 
ll sum(ll a, ll b) {ll res = (a + b) % mod;if (res < 0) res += mod;return res;}
ll mult(ll a, ll b) {ll res = a * b % mod;if (res < 0) res += mod;return res;}
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
typedef vector<vpll> vvpll;
void    test_case()
{
    //must read the constraints properly
    int  n,k;
    cin>>k;
    n=(1<<k)-1;
    int cnt=1;
    vector<int>dp(n+1,1);
    string s;
    cin>>s;
    s='&'+s;
    // cin>>s;
    umap<int,int>child,parent;
    for(int i=1;i<=(1<<(k-1));i++){
        child[i]=0;
        if(s[i]=='?')
        dp[i]=2;
    }
    // debug(s);
    // debug((1<<k-1)+1,n);
    for(int i=(1<<k-1)+1;i<=n;i++){
        child[i]=cnt;
        parent[cnt+1]=parent[cnt]=i;
        cnt+=2;
        if(s[i]=='0')
        dp[i]=dp[child[i]];
        else if(s[i]=='1')
        dp[i]=dp[child[i]+1];
        else dp[i]=dp[child[i]]+dp[child[i]+1];
    }
    parent[n]=n;
    int ans=dp[(1<<k)-1];
    int q;
    cin>>q;
    // debug(parent,child);
    // debug(dp);
    while(q--){
        int x;
        cin>>x;
        cin>>s[x];
        int prev=-1;
        while(prev!=x){
            prev=x;
            if(child[x]==0){
                dp[x]=s[x]=='?'?2:1;
                x=parent[x];
                continue;
            }
            if(s[x]=='0')
                dp[x]=dp[child[x]];
            else if(s[x]=='1')
                dp[x]=dp[child[x]+1];
            else dp[x]=dp[child[x]]+dp[child[x]+1];
            x=parent[x];
        }
        cout<<dp[n]<<endl;
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    fast;
    ll t=1;
    // cin>>t;
    while(t--){
          test_case();
    }
    return 0;
    
}