Problem: Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a0, a1, ..., ah, where h is the height of the tree, and ai equals to the number of vertices that are at distance of i edges from root. 

Unfortunately, this time Sasha's intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence ai, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.

Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.

The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.

Input

The first line contains a single integer h (2 ≤ h ≤ 105) — the height of the tree.

The second line contains h + 1 integers — the sequence a0, a1, ..., ah (1 ≤ ai ≤ 2·105). The sum of all ai does not exceed 2·105. It is guaranteed that there is at least one tree matching this sequence.

Output

If there is only one tree matching this sequence, print "perfect".

Otherwise print "ambiguous" in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print <image> integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.

These treese should be non-isomorphic and should match the given sequence.

Examples

Input

2
1 1 1


Output

perfect


Input

2
1 2 2


Output

ambiguous
0 1 1 3 3
0 1 1 3 2

Note

The only tree in the first example and the two printed trees from the second example are shown on the picture:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
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 pr[2000010];
long long inv[2000010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1ll;
  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() {
  pr[0] = 1;
  for (long long i = 1; i < 2000010; i++) {
    pr[i] = (pr[i - 1] * i) % 1000000007;
  }
  for (long long i = 0; i < 2000010; i++)
    inv[i] = beki(pr[i], 1000000007 - 2, 1000000007);
}
long long par[200100], ranks[200100], kosuu[200100];
void shoki(long long n) {
  for (long long i = 0; i < n; i++) {
    par[i] = i;
    ranks[i] = 0;
    kosuu[i] = 1;
  }
}
long long root(long long x) { return par[x] == x ? x : par[x] = root(par[x]); }
bool same(long long x, long long y) { return root(x) == root(y); }
void unite(long long x, long long y) {
  x = root(x);
  y = root(y);
  long long xx = kosuu[x], yy = kosuu[y];
  if (x == y) return;
  if (ranks[x] < ranks[y]) {
    par[x] = y;
    kosuu[y] = yy + xx;
  } else {
    par[y] = x;
    if (ranks[x] == ranks[y]) ranks[x] = ranks[x] + 1;
    kosuu[x] = yy + xx;
  }
  return;
}
long long a[100030];
long long rui[100030] = {0};
vector<long long> ve, ve2;
signed main() {
  long long n, m;
  cin >> n;
  for (long long i = 0; i <= n; i++) cin >> a[i];
  bool bo = true;
  for (long long i = 1; i <= n; i++) {
    if (a[i] > 1 && a[i - 1] > 1) {
      bo = 0;
    }
  }
  if (bo) {
    cout << "perfect" << endl;
    return 0;
  }
  rui[0] = a[0];
  for (long long i = 1; i <= n; i++) rui[i] = rui[i - 1] + a[i];
  ve.push_back(0);
  ve2.push_back(0);
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j < a[i]; j++) {
      ve.push_back(rui[i - 1]);
      ve2.push_back(rui[i - 1]);
    }
    if (a[i] > 1 && a[i - 1] > 1) {
      long long u = ve2.back();
      ve2.back() = u - 1;
    }
  }
  cout << "ambiguous" << endl;
  for (auto v : ve) cout << v << " ";
  cout << endl;
  for (auto v : ve2) cout << v << " ";
  cout << endl;
  return 0;
}