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:
#pragma optimize("SEX_ON_THE_BEACH")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("unroll-all-loops")
#pragma GCC optimize("O3")
 
#pragma GCC optimize("Ofast")
#pragma GCC optimize("fast-math")
#define _FORTIFY_SOURCE 0 
#pragma GCC optimize("no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native") 
 
#include<bits/stdc++.h>
 
using ll = long long int;
using ull = unsigned long long int;
using dd = double;
using ldd = long double;
 
namespace someUsefull {
    template<typename T1, typename T2>
    inline void checkMin(T1& a, T2 b) {
        if (a > b)
            a = b;
    }
 
    template<typename T1, typename T2>
    inline void checkMax(T1& a, T2 b) {
        if (a < b)
            a = b;
    }
 
    const int _bitfunctions_size_ = 16;
    const int _bitfunctions_mask_ = (1 << _bitfunctions_size_) - 1;
    char* _bits_count_;
    
    inline void _build_bits_count_() {
        _bits_count_ = new char[1 << _bitfunctions_size_];
        for (int i = 1; i < (1 << _bitfunctions_size_); ++i) {
            _bits_count_[i] = _bits_count_[i >> 1];
            if (i & 1) ++_bits_count_[i];
        }
    }
 
    template<typename T1 = char, typename T2 = int>
    inline T1 popcnt(T2 x) {
        T1 ans = 0;
        while (x) {
            ans += _bits_count_[x & _bitfunctions_mask_];
            x >>= _bitfunctions_size_;
        }
        return ans;
    }
}
 
namespace operators {
    template<typename T1, typename T2>
    std::istream& operator>>(std::istream& in, std::pair<T1, T2>& x) {
        in >> x.first >> x.second;
        return in;
    }
 
    template<typename T1, typename T2>
    std::ostream& operator<<(std::ostream& out, std::pair<T1, T2> x) {
        out << x.first << " " << x.second;
        return out;
    }
 
    template<typename T1>
    std::istream& operator>>(std::istream& in, std::vector<T1>& x) {
        for (auto& i : x) in >> i;
        return in;
    }
 
    template<typename T1>
    std::ostream& operator<<(std::ostream& out, std::vector<T1>& x) {
        for (auto& i : x) out << i << " ";
        return out;
    }
}
 
//name spaces
using namespace std;
using namespace operators;
using namespace someUsefull;
//end of name spaces
 
//defines
#define ff first
#define ss second
#define all(x) (x).begin(), (x).end()
 
 
//end of defines
//#undef HOME
//debug defines
#ifdef HOME
    #define debug(x) cerr << #x << " " << x << endl;
    #define debug_v(x) {cerr << #x << " "; for (auto ioi : x) cerr << ioi << " "; cerr << endl;}
    #define debug_vp(x) {cerr << #x << " "; for (auto ioi : x) cerr << '[' << ioi.ff << " " << ioi.ss << ']'; cerr << endl;}
    #define debug_v_v(x) {cerr << #x << "/*\n"; for (auto ioi : x) { for (auto ioi2 : ioi) cerr << ioi2 << " "; cerr << '\n';} cerr << "*/" << #x << endl;}
    int jjj;
    #define wait() cin >> jjj;
    #define PO cerr << "POMELO" << endl;
    #define OL cerr << "OLIVA" << endl;
    #define gen_clock(x) cerr << "Clock " << #x << " created" << endl; ll x = clock(); 
    #define check_clock(x) cerr << "Time spent in " << #x << ": " << clock() - x << endl; x = clock();
    #define reset_clock(x) x = clock();
    #define say(x) cerr << x << endl;
#else
    #define debug(x) 0;
    #define debug_v(x) 0;
    #define debug_vp(x) 0;
    #define debug_v_v(x) 0;
    #define wait() 0;
    #define PO 0;
    #define OL 0;
    #define gen_clock(x) 0;
    #define check_clock(x) 0;
    #define reset_clock(x) 0;
    #define say(x) 0;
#endif // HOME

vector<char> tree;
vector<int> dp;
int _size;

void solve(int test) {
    int k;
    cin >> k;
    tree.resize(1 << k, '_');
    dp.resize(2 << k, 1);
    _size = 1 << k;
    for (int i = tree.size() - 1; i > 0; --i) {
        cin >> tree[i];
    }
    for (int i = _size - 1; i > 0; --i) {
        if (tree[i] == '?') {
            dp[i] = dp[i << 1] + dp[i << 1 | 1];
        } else if (tree[i] == '0') {
            dp[i] = dp[i << 1 | 1];
        } else {
            dp[i] = dp[i << 1];
        }
    }
    int q;
    cin >> q;
    for (int i = 0; i < q; ++i) {
        int x;
        cin >> x;
        char z ;
        cin >> z;
        x = _size - x;
        tree[x] = z;
        while (x) {
            if (tree[x] == '?') {
                dp[x] = dp[x << 1] + dp[x << 1 | 1];
            } else if (tree[x] == '0') {
                dp[x] = dp[x << 1 | 1];
            } else {
                dp[x] = dp[x << 1];
            }
            x >>= 1;
        }
        debug_v(tree);
        debug_v(dp);
        cout << dp[1] << '\n';
        PO;
    }
}
 
signed main() {
    ios_base::sync_with_stdio(false);
    cout.tie(0);
    cin.tie(0);
    _build_bits_count_();
    int t = 1;
    //cin >> t;
    for (int i = 0; i < t; ++i) {
        solve(i+1);
    }
    return 0;
}
/*
 
*/