Problem: Allen and Bessie are playing a simple number game. They both know a function f: \{0, 1\}^n → R, i. e. the function takes n binary arguments and returns a real value. At the start of the game, the variables x_1, x_2, ..., x_n are all set to -1. Each round, with equal probability, one of Allen or Bessie gets to make a move. A move consists of picking an i such that x_i = -1 and either setting x_i → 0 or x_i → 1.

After n rounds all variables are set, and the game value resolves to f(x_1, x_2, ..., x_n). Allen wants to maximize the game value, and Bessie wants to minimize it.

Your goal is to help Allen and Bessie find the expected game value! They will play r+1 times though, so between each game, exactly one value of f changes. In other words, between rounds i and i+1 for 1 ≤ i ≤ r, f(z_1, ..., z_n) → g_i for some (z_1, ..., z_n) ∈ \{0, 1\}^n. You are to find the expected game value in the beginning and after each change.

Input

The first line contains two integers n and r (1 ≤ n ≤ 18, 0 ≤ r ≤ 2^{18}).

The next line contains 2^n integers c_0, c_1, ..., c_{2^n-1} (0 ≤ c_i ≤ 10^9), denoting the initial values of f. More specifically, f(x_0, x_1, ..., x_{n-1}) = c_x, if x = \overline{x_{n-1} … x_0} in binary.

Each of the next r lines contains two integers z and g (0 ≤ z ≤ 2^n - 1, 0 ≤ g ≤ 10^9). If z = \overline{z_{n-1} ... z_0} in binary, then this means to set f(z_0, ..., z_{n-1}) → g.

Output

Print r+1 lines, the i-th of which denotes the value of the game f during the i-th round. Your answer must have absolute or relative error within 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Examples

Input

2 2
0 1 2 3
2 5
0 4


Output

1.500000
2.250000
3.250000


Input

1 0
2 3


Output

2.500000


Input

2 0
1 1 1 1


Output

1.000000

Note

Consider the second test case. If Allen goes first, he will set x_1 → 1, so the final value will be 3. If Bessie goes first, then she will set x_1 → 0 so the final value will be 2. Thus the answer is 2.5.

In the third test case, the game value will always be 1 regardless of Allen and Bessie's play.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
long long dx[4] = {0, -1, 0, 1};
long long dy[4] = {1, 0, -1, 0};
using namespace std;
class pa3 {
 public:
  long long x, y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  double x;
  long long y, z, w;
  pa4(double x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
class Point {
 public:
  double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
  Point operator+(Point p) { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) { return Point(x - p.x, y - p.y); }
  Point operator*(double a) { return Point(x * a, y * a); }
  Point operator/(double a) { return Point(x / a, y / a); }
  double absv() { return sqrt(norm()); }
  double norm() { return x * x + y * y; }
  bool operator<(const Point &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const Point &p) const {
    return fabs(x - p.x) < (1e-10) && fabs(y - p.y) < (1e-10);
  }
};
struct Segment {
  Point p1, p2;
};
double dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
bool parareru(Point a, Point b, Point c, Point d) {
  return abs(cross(a - b, d - c)) < (1e-10);
}
double distance_ls_p(Point a, Point b, Point c) {
  if (dot(b - a, c - a) < (1e-10)) return (c - a).absv();
  if (dot(a - b, c - b) < (1e-10)) return (c - b).absv();
  return abs(cross(b - a, c - a)) / (b - a).absv();
}
bool is_intersected_ls(Segment a, Segment b) {
  if (a.p1 == b.p1 || a.p2 == b.p1 || a.p1 == b.p2 || a.p2 == b.p2)
    return false;
  if (parareru((a.p2), (a.p1), (a.p1), (b.p2)) &&
      parareru((a.p2), (a.p1), (a.p1), (b.p1))) {
    if (dot(a.p1 - b.p1, a.p1 - b.p2) < (1e-10)) return true;
    if (dot(a.p2 - b.p1, a.p2 - b.p2) < (1e-10)) return true;
    if (dot(a.p1 - b.p1, a.p2 - b.p1) < (1e-10)) return true;
    if (dot(a.p1 - b.p2, a.p2 - b.p2) < (1e-10)) return true;
    return false;
  } else
    return (cross(a.p2 - a.p1, b.p1 - a.p1) * cross(a.p2 - a.p1, b.p2 - a.p1) <
            (1e-10)) &&
           (cross(b.p2 - b.p1, a.p1 - b.p1) * cross(b.p2 - b.p1, a.p2 - b.p1) <
            (1e-10));
}
double segment_dis(Segment a, Segment b) {
  if (is_intersected_ls(a, b)) return 0;
  double r = distance_ls_p(a.p1, a.p2, b.p1);
  r = min(r, distance_ls_p(a.p1, a.p2, b.p2));
  r = min(r, distance_ls_p(b.p1, b.p2, a.p2));
  r = min(r, distance_ls_p(b.p1, b.p2, a.p1));
  return r;
}
Point intersection_ls(Segment a, Segment b) {
  Point ba = b.p2 - b.p1;
  double d1 = abs(cross(ba, a.p1 - b.p1));
  double d2 = abs(cross(ba, a.p2 - b.p1));
  double t = d1 / (d1 + d2);
  return a.p1 + (a.p2 - a.p1) * t;
}
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long gcd(long long v, long long b) {
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
double distans(double x1, double y1, double x2, double y2) {
  double rr = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
  return sqrt(rr);
}
long long mod;
long long pr[2010];
long long inv[2010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  if (rr % 2 == 1) return (beki(wa, rr - 1, warukazu) * wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i < ert; i++) {
    pr[i] = (pr[i - 1] * i) % mod;
  }
  for (long long i = 0; i < ert; i++) inv[i] = beki(pr[i], mod - 2, mod);
}
long long a[1 << 18];
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, r;
  cin >> n >> r;
  long long ans = 0;
  for (long long i = 0; i < (1 << n); i++) {
    cin >> a[i];
    ans += a[i];
  }
  printf("%.10lf\n", (double)ans / ((1 << n) + 0.0));
  for (long long i = 0; i < r; i++) {
    long long x, e;
    cin >> x >> e;
    long long sa = e - a[x];
    a[x] = e;
    ans += sa;
    printf("%.10lf\n", (double)ans / ((1 << n) + 0.0));
  }
  return 0;
}