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;
using namespace __detail;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
#define ar array
#define vi vector<int>
#define vvi vector<vector<int>>
#define vll vector<long long>
#define all(x) (x).begin(), (x).end()
#define en '\n'
#define FILL(x, v) memset(x, v, sizeof(x))
#define pb push_back

#define fast                   \
  ios::sync_with_stdio(false); \
  cin.tie(NULL);               \
  cout.tie(NULL);
#define F_OR(i, a, b, s) \
  for (int i = (a); ((s) > 0 ? i < (b) : i > (b)); i += (s))
#define F_OR1(e) F_OR(i, 0, e, 1)
#define F_OR2(i, e) F_OR(i, 0, e, 1)
#define F_OR3(i, b, e) F_OR(i, b, e, 1)
#define F_OR4(i, b, e, s) F_OR(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)
#define FOR(...)     \
  F_ORC(__VA_ARGS__) \
  (__VA_ARGS__)

#define E_ACH2(x, a) for (auto &x : a)
#define E_ACH3(x, y, a) for (auto &[x, y] : a)
#define E_ACH4(x, y, z, a) for (auto &[x, y, z] : a)
#define E_ACHC(...) GET5(__VA_ARGS__, E_ACH4, E_ACH3, E_ACH2)
#define EACH(...)     \
  E_ACHC(__VA_ARGS__) \
  (__VA_ARGS__)

#define sim template <class c
#define ris return *this
#define dor > debug &operator<<
#define eni(x)                                                              \
  sim > typename enable_if<sizeof dud<c>(0) x 1, debug &>::type operator<<( \
            c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c *x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
  ~debug() { cerr << endl; }
  eni(!=) cerr << boolalpha << i;
  ris;
} eni(==) ris << range(begin(i), end(i));
}
sim, class b dor(pair<b, c> d) {
  ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
  *this << "[";
  for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
  ris << "]";
}
#else
  sim dor(const c &) { ris; }
#endif
}
;
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "
// debug & operator << (debug & dd, P p) { dd << "(" << p.x << ", " << p.y <<
// ")"; return dd; }

template <typename A, typename B>
inline void chmin(A &a, B b) {
  if (a > b) a = b;
}
template <typename A, typename B>
inline void chmax(A &a, B b) {
  if (a < b) a = b;
}

const int M = 1e9 + 7;

int mul(const int v1, const int v2) { return (int)((1LL * v1 * v2) % M); }

int mod_pow(int a, int p) {
  int res = 1;
  while (p != 0) {
    if (p & 1) res = mul(res, a);
    p >>= 1;
    a = mul(a, a);
  }
  return res;
}

/*
DEBUG: -D LOCAL
debug() << imie(k) imie(x) imie(row) imie(col);
*/

const int N = (1 << 18);
int tree[4 * N];
string s;

void countPossibleWinners(int i) {
  tree[i] = 0;
  if (s[i] != '1') {
    tree[i] += tree[i * 2 + 2];  // + right
  }
  if (s[i] != '0') {
    tree[i] += tree[i * 2 + 1];  // + left;
  }
  // debug() << imie(tree[i]);
}

/**
 * i -> current node
 * l, r -> left and rightmost range from current node
 */
void build(int i, int l, int r) {
  if (l == r) {  // leaf nodes
    if (s[i] == '?') {
      tree[i] = 2;
    } else {
      tree[i] = 1;
    }
    return;
  }

  int m = (l + r) / 2;
  build(2 * i + 1, l, m);
  build(2 * i + 2, m + 1, r);
  // debug() << imie(l) imie(m) imie(r);
  countPossibleWinners(i);
}

int main() {
  fast;
  int k, q;
  cin >> k >> s >> q;

  fill(tree, tree + 4 * N, 1);
  reverse(all(s));
  build(0, 1, (1 << (k - 1)));

  while (q--) {
    int p;
    char c;
    cin >> p >> c;

    int i = (1 << k) - 1 - p;
    s[i] = c;

    while (1) {
      countPossibleWinners(i);
      if (i == 0) {
        break;
      }
      i = (i - 1) / 2;
    }

    cout << tree[0] << en;
  }
}