#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("-ffloat-store")
using namespace std;
clock_t time_p = clock();
void aryanc403() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || a.first == b.first && a.second <= b.second);
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 1000000007L;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m;
string s;
vector<long long int> a, b;
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    cin >> n >> m;
    a.clear();
    a.reserve(n);
    b.resize(n + 1, 0);
    for (i = 0; i < (m); ++i) {
      cin >> in;
      a.push_back(in);
    }
    for (i = 0; i < (m - 1); ++i) {
      long long int x = a[i + 1] - a[i];
      x %= n;
      x = (x + n) % n;
      if (x == 0) x += n;
      b[a[i]] = x;
    }
    for (i = (1); i <= (n); ++i) {
      if (b[i]) continue;
      for (j = (1); j <= (n); ++j) {
        bool fl = true;
        for (auto x : b) {
          if (x == j) {
            fl = false;
            break;
          }
        }
        if (fl) {
          b[i] = j;
          break;
        }
      }
    };
    for (i = 0; i < (m - 1); ++i) {
      long long int x = a[i + 1] - a[i];
      x %= n;
      x = (x + n) % n;
      if (x == 0) x += n;
      if (b[a[i]] != x) {
        cout << -1 << "\n";
        return 0;
      }
    }
    a = b;
    sort(a.begin(), a.end());
    a.erase(unique(a.begin(), a.end()), a.end());
    if ((long long int)a.size() <= n) {
      cout << -1 << "\n";
      return 0;
    }
    for (i = (1); i <= (n); ++i) cout << b[i] << " ";
  }
  aryanc403();
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

n children are standing in a circle and playing a game. Children's numbers in clockwise order form a permutation a1, a2, ..., an of length n. It is an integer sequence such that each integer from 1 to n appears exactly once in it.

The game consists of m steps. On each step the current leader with index i counts out ai people in clockwise order, starting from the next person. The last one to be pointed at by the leader becomes the new leader.

You are given numbers l1, l2, ..., lm — indices of leaders in the beginning of each step. Child with number l1 is the first leader in the game. 

Write a program which will restore a possible permutation a1, a2, ..., an. If there are multiple solutions then print any of them. If there is no solution then print -1.

Input

The first line contains two integer numbers n, m (1 ≤ n, m ≤ 100).

The second line contains m integer numbers l1, l2, ..., lm (1 ≤ li ≤ n) — indices of leaders in the beginning of each step.

Output

Print such permutation of n numbers a1, a2, ..., an that leaders in the game will be exactly l1, l2, ..., lm if all the rules are followed. If there are multiple solutions print any of them. 

If there is no permutation which satisfies all described conditions print -1.

Examples

Input

4 5
2 3 1 4 4


Output

3 1 2 4 


Input

3 3
3 1 2


Output

-1

Note

Let's follow leadership in the first example: 

  * Child 2 starts. 
  * Leadership goes from 2 to 2 + a2 = 3. 
  * Leadership goes from 3 to 3 + a3 = 5. As it's greater than 4, it's going in a circle to 1. 
  * Leadership goes from 1 to 1 + a1 = 4. 
  * Leadership goes from 4 to 4 + a4 = 8. Thus in circle it still remains at 4. 

### Response

```cpp
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("-ffloat-store")
using namespace std;
clock_t time_p = clock();
void aryanc403() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
class CMP {
 public:
  bool operator()(pair<long long int, long long int> a,
                  pair<long long int, long long int> b) {
    return !(a.first < b.first || a.first == b.first && a.second <= b.second);
  }
};
void add(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt == m.end())
    m.insert({x, cnt});
  else
    jt->second += cnt;
}
void del(map<long long int, long long int> &m, long long int x,
         long long int cnt = 1) {
  auto jt = m.find(x);
  if (jt->second <= cnt)
    m.erase(jt);
  else
    jt->second -= cnt;
}
bool cmp(const pair<long long int, long long int> &a,
         const pair<long long int, long long int> &b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
const long long int mod = 1000000007L;
long long int T, n, i, j, k, in, cnt, l, r, u, v, x, y;
long long int m;
string s;
vector<long long int> a, b;
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    cin >> n >> m;
    a.clear();
    a.reserve(n);
    b.resize(n + 1, 0);
    for (i = 0; i < (m); ++i) {
      cin >> in;
      a.push_back(in);
    }
    for (i = 0; i < (m - 1); ++i) {
      long long int x = a[i + 1] - a[i];
      x %= n;
      x = (x + n) % n;
      if (x == 0) x += n;
      b[a[i]] = x;
    }
    for (i = (1); i <= (n); ++i) {
      if (b[i]) continue;
      for (j = (1); j <= (n); ++j) {
        bool fl = true;
        for (auto x : b) {
          if (x == j) {
            fl = false;
            break;
          }
        }
        if (fl) {
          b[i] = j;
          break;
        }
      }
    };
    for (i = 0; i < (m - 1); ++i) {
      long long int x = a[i + 1] - a[i];
      x %= n;
      x = (x + n) % n;
      if (x == 0) x += n;
      if (b[a[i]] != x) {
        cout << -1 << "\n";
        return 0;
      }
    }
    a = b;
    sort(a.begin(), a.end());
    a.erase(unique(a.begin(), a.end()), a.end());
    if ((long long int)a.size() <= n) {
      cout << -1 << "\n";
      return 0;
    }
    for (i = (1); i <= (n); ++i) cout << b[i] << " ";
  }
  aryanc403();
  return 0;
}

```