Problem: Petya had a tree consisting of n vertices numbered with integers from 1 to n. Accidentally he lost his tree. 

Petya remembers information about k vertices: distances from each of them to each of the n tree vertices.

Your task is to restore any tree that satisfies the information that Petya remembers or report that such tree doesn't exist.

Input

The first line contains two integers n and k (2 ≤ n ≤ 30 000, 1 ≤ k ≤ min(200, n)) — the number of vertices in the tree and the number of vertices about which Petya remembers distance information.

The following k lines contain remembered information. The i-th line contains n integers di, 1, di, 2, ..., di, n (0 ≤ di, j ≤ n - 1), where di, j — the distance to j-th vertex from the i-th vertex that Petya remembers.

Output

If there are no suitable trees, print -1.

In the other case, print n - 1 lines: each line should contain two vertices connected by edge in the required tree. You can print edges and vertices in an edge in any order. The tree vertices are enumerated from 1 to n.

If there are many solutions print any of them.

Examples

Input

5 2
0 1 2 3 2
2 1 0 1 2


Output

2 1
3 2
4 3
5 2


Input

3 1
1 2 1


Output

-1

Note

Picture for the first sample:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1000000007;
const char nl = '\n';
const int MX = 100001;
int base;
struct hsh {
  long long base, p1, p2;
  long long val1, val2;
  long long inv1, inv2;
  long long modExp(long long power, long long prime) {
    if (power == 0) {
      return 1;
    } else {
      long long cur = modExp(power / 2, prime);
      cur = cur * cur;
      cur = cur % prime;
      if (power % 2 == 1) cur = cur * base;
      cur = cur % prime;
      return cur;
    }
  }
  hsh(int b) {
    base = b;
    p1 = 1000000007;
    p2 = 1000000009;
    val1 = 0;
    val2 = 0;
    inv2 = modExp(p2 - 2, p2);
    inv1 = modExp(p1 - 2, p1);
  }
  hsh(long long b, string S) {
    p1 = 1000000007;
    p2 = 1000000009;
    base = b;
    val1 = 0;
    val2 = 0;
    inv2 = modExp(p2 - 2, p2);
    inv1 = modExp(p1 - 2, p1);
    for (int i = 0; i < ((int)(S).size()); i++) {
      push_back(S[i] - 'a');
    }
  }
  void push_back(long long v) {
    val1 *= base;
    val1 %= p1;
    val1 += v;
    val1 %= p1;
    val2 *= base;
    val2 %= p2;
    val2 += v;
    val2 %= p2;
  }
  pair<long long, long long> get() { return {val1, val2}; }
};
void prepHash() { base = uniform_int_distribution<int>(1000, MOD - 2)(rng); }
template <int SZ>
struct DSU {
  int parent[SZ], size[SZ];
  DSU() {
    for (int i = 0; i < (SZ); i++) parent[i] = i, size[i] = 0;
  }
  int get(int x) {
    if (parent[x] != x) parent[x] = get(parent[x]);
    return parent[x];
  }
  void unify(int x, int y) {
    x = get(x);
    y = get(y);
    if (x == y) return;
    if (size[x] < size[y]) swap(x, y);
    if (size[x] == size[y]) size[x]++;
    parent[y] = x;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  prepHash();
  int N, K;
  cin >> N >> K;
  int D[K][N];
  int V[K];
  bool used[N];
  for (int i = 0; i < (N); i++) used[i] = false;
  for (int i = 0; i < (K); i++) {
    V[i] = -1;
    for (int j = 0; j < (N); j++) {
      cin >> D[i][j];
      if (D[i][j] == 0) {
        if (V[i] != -1) {
          cout << -1 << nl;
          return 0;
        }
        V[i] = j;
      }
    }
    if (V[i] == -1) {
      cout << -1 << nl;
      return 0;
    }
    if (used[V[i]]) {
      cout << -1 << nl;
      return 0;
    }
    used[V[i]] = true;
  }
  vector<vector<int> > graph(N);
  bool proc[N];
  for (int i = 0; i < (N); i++) proc[i] = false;
  proc[V[0]] = true;
  pair<long long, long long> H[N];
  set<pair<pair<long long, long long>, int> > hmap;
  hmap.insert({{2000000000, 2000000000}, -1});
  for (int i = 0; i < (N); i++) {
    hsh cur(base);
    for (int j = 0; j < (K); j++) {
      cur.push_back(D[j][i]);
    }
    H[i] = cur.get();
    hmap.insert({H[i], i});
  }
  hmap.erase({H[V[0]], V[0]});
  queue<int> q;
  q.push(V[0]);
  for (int i = 1; i < (K); i++) {
    vector<int> line(D[0][V[i]] + 1);
    for (int j = 0; j < (D[0][V[i]] + 1); j++) line[j] = -1;
    line[D[0][V[i]]] = V[i];
    line[0] = V[0];
    for (int j = 0; j < (N); j++) {
      if (D[0][j] + D[i][j] == D[0][V[i]]) {
        line[D[0][j]] = j;
      }
    }
    for (auto& a : line) {
      if (a == -1) {
        cout << -1 << nl;
        return 0;
      }
    }
    for (int i = 1; i < ((int)(line).size()); i++) {
      if (!proc[line[i]]) {
        proc[line[i]] = true;
        graph[line[i]].push_back(line[i - 1]);
        graph[line[i - 1]].push_back(line[i]);
        hmap.erase({H[line[i]], line[i]});
        q.push(line[i]);
      }
    }
  }
  hsh oneCur(base);
  for (int i = 0; i < (K); i++) oneCur.push_back(1);
  pair<long long, long long> oneHash = oneCur.get();
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    pair<long long, long long> curHash = {H[cur].first + oneHash.first,
                                          H[cur].second + oneHash.second};
    curHash.first %= MOD;
    curHash.second %= MOD + 2;
    while (hmap.lower_bound({curHash, -1})->first == curHash) {
      int v = hmap.lower_bound({curHash, -1})->second;
      hmap.erase(hmap.lower_bound({curHash, -1}));
      q.push(v);
      graph[cur].push_back(v);
      graph[v].push_back(cur);
    }
  }
  hmap.erase(hmap.begin());
  if (!hmap.empty()) {
    cout << -1 << nl;
    return 0;
  }
  int dist[K][N];
  for (int r = 0; r < (K); r++) {
    for (int i = 0; i < (N); i++) {
      dist[r][i] = N + 1;
    }
    dist[r][V[r]] = 0;
    q.push(V[r]);
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      for (auto& a : graph[cur]) {
        if (dist[r][a] > dist[r][cur] + 1) {
          dist[r][a] = dist[r][cur] + 1;
          q.push(a);
        }
      }
    }
  }
  for (int i = 0; i < (K); i++) {
    for (int j = 0; j < (N); j++) {
      if (dist[i][j] != D[i][j]) {
        cout << -1 << nl;
        return 0;
      }
    }
  }
  for (int i = 0; i < (N); i++) {
    for (auto& a : graph[i]) {
      if (a > i) continue;
      cout << i + 1 << " " << a + 1 << nl;
    }
  }
  return 0;
}