Andrey received a postcard from Irina. It contained only the words "Hello, Andrey!", and a strange string consisting of lowercase Latin letters, snowflakes and candy canes. Andrey thought that this string is an encrypted message, and decided to decrypt it.

Andrey noticed that snowflakes and candy canes always stand after the letters, so he supposed that the message was encrypted as follows. Candy cane means that the letter before it can be removed, or can be left. A snowflake means that the letter before it can be removed, left, or repeated several times.

For example, consider the following string: 

<image>

This string can encode the message «happynewyear». For this, candy canes and snowflakes should be used as follows: 

  * candy cane 1: remove the letter w, 
  * snowflake 1: repeat the letter p twice, 
  * candy cane 2: leave the letter n, 
  * snowflake 2: remove the letter w, 
  * snowflake 3: leave the letter e. 

<image>

Please note that the same string can encode different messages. For example, the string above can encode «hayewyar», «happpppynewwwwwyear», and other messages.

Andrey knows that messages from Irina usually have a length of k letters. Help him to find out if a given string can encode a message of k letters, and if so, give an example of such a message.

Input

The first line contains the string received in the postcard. The string consists only of lowercase Latin letters, as well as the characters «*» and «?», meaning snowflake and candy cone, respectively. These characters can only appear immediately after the letter. The length of the string does not exceed 200.

The second line contains an integer number k (1 ≤ k ≤ 200), the required message length.

Output

Print any message of length k that the given string can encode, or «Impossible» if such a message does not exist.

Examples

Input


hw?ap*yn?eww*ye*ar
12


Output


happynewyear


Input


ab?a
2


Output


aa

Input


ab?a
3


Output


aba

Input


ababb
5


Output


ababb

Input


ab?a
1


Output


Impossible
#include <bits/stdc++.h>
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  long long x;
  long long 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:
  long long x;
  long long y, z, w;
  pa4(long long 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);
  }
};
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 extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first + r.first, l.second + r.second};
}
long long pr[1200010];
long long inv[1200010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
double bekid(double w, long long r) {
  if (r == 0) return 1.0;
  if (r == 1) return w;
  if (r % 2) return bekid(w, r - 1) * w;
  double f = bekid(w, r / 2);
  return f * f;
}
long long comb(long long nn, long long rr) {
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
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);
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  long long t;
  cin >> t;
  long long ch = 0;
  long long yu = 0;
  long long ky = 0;
  for (auto c : s) {
    if (c != '*' && c != '?')
      ch++;
    else if (c == '*')
      yu++;
    else
      ky++;
  }
  if (yu + ky == 0) {
    if (t == ch)
      cout << s << endl;
    else
      cout << "Impossible" << endl;
    return 0;
  }
  if (yu == 0) {
    if (!(ch - ky <= t && t <= ch))
      cout << "Impossible" << endl;
    else {
      string ans = "";
      long long nozo = ch - t;
      for (auto c : s) {
        if (c != '*' && c != '?')
          ans += c;
        else if (nozo)
          ans.pop_back(), nozo--;
      }
      cout << ans << endl;
    }
    return 0;
  }
  {
    if (ch - ky - yu > t) {
      cout << "Impossible" << endl;
      return 0;
    }
    string ans = "";
    long long e = 0;
    char r;
    for (auto c : s) {
      if (c != '*' && c != '?')
        ans += c;
      else if (c == '?')
        ans.pop_back();
      else if (c == '*' && e == 0) {
        e++;
        r = ans.back();
        ans.pop_back();
        ans += "*";
      } else {
        ans.pop_back();
      }
    }
    long long tasu = t - ans.length() + 1;
    for (auto c : ans) {
      if (c == '*') {
        for (long long i = 0; i < tasu; i++) cout << r;
      } else
        cout << c;
    }
    cout << endl;
  }
  return 0;
}
