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:

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const ll MOD = 1000000007;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
ll binpow(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
ll modInv(ll a) { return binpow(a, MOD - 2); }
const double PI = acos(-1);
const double eps = -1e6;
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
const int mxN = 300001;
int par[mxN][16];
int depth[mxN];
vector<int> spe[201];
int val[201];
vector<int> hang[mxN];
vector<pair<int, vector<pair<int, int>>>> rst;
vector<pair<int, int>> res;
int cnt[mxN];
bool flag[mxN];
int n, k;
void killthyself() {
  cout << "-1\n";
  exit(0);
}
void upd(int a, int p) {
  if (flag[a]) return;
  flag[a] = true;
  par[a][0] = p;
  depth[a] = depth[p] + 1;
  for (int lg = (1); lg < (16); ++lg) {
    par[a][lg] = par[par[a][lg - 1]][lg - 1];
  }
  res.push_back({a, p});
}
int walk(int a, int nh) {
  for (int i = (16) - 1; i >= (0); --i) {
    if (depth[par[a][i]] >= nh) {
      a = par[a][i];
    }
  }
  return a;
}
int lca(int a, int b) {
  if (depth[a] < depth[b]) {
    b = walk(b, depth[a]);
  } else if (depth[a] > depth[b]) {
    a = walk(a, depth[b]);
  }
  if (a == b) return a;
  for (int i = (16) - 1; i >= (0); --i) {
    if (par[a][i] != par[b][i]) {
      a = par[a][i];
      b = par[b][i];
    }
  }
  return par[a][0];
}
int dist(int a, int b) { return depth[a] + depth[b] - 2 * depth[lca(a, b)]; }
int nthonpath(int first, int second, int d) {
  int imhavingfun = lca(first, second);
  if (depth[first] - depth[imhavingfun] >= d) {
    return walk(first, depth[first] - d);
  } else {
    d -= (depth[first] - depth[imhavingfun]);
    return walk(second, depth[imhavingfun] + d);
  }
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  cin.tie(0)->sync_with_stdio(0);
  cin >> n >> k;
  for (int i = (0); i < (n); ++i) {
    hang[i].push_back(i);
  }
  for (int i = (0); i < (k); ++i) {
    memset(cnt, 0, sizeof(cnt));
    spe[i].resize(n);
    int mx = 0;
    for (int j = (0); j < (n); ++j) {
      cin >> spe[i][j];
      if (spe[i][j] == 0) val[i] = j;
      mx = max(mx, spe[i][j]);
      cnt[spe[i][j]]++;
    }
    int mex = -1;
    while (cnt[mex + 1]) mex++;
    if (mex != mx) killthyself();
  }
  int rt = val[0];
  for (int i = (0); i < (16); ++i) {
    par[rt][i] = rt;
  }
  flag[rt] = 1;
  for (int i = (1); i < (k); ++i) {
    vector<pair<int, int>> chain;
    for (int j = (0); j < (n); ++j) {
      if (spe[i][j] + spe[0][j] == spe[0][val[i]])
        chain.push_back({spe[0][j], j});
    }
    sort(chain.begin(), chain.end());
    int prv = -1;
    int tp = rt;
    for (int j = (0); j < (chain.size()); ++j) {
      if (chain[j].first != prv + 1) killthyself();
      prv++;
      upd(chain[j].second, tp);
      tp = chain[j].second;
    }
  }
  for (int i = (0); i < (n); ++i) {
    if (!flag[i]) {
      rst.push_back({i, {}});
      for (int j = (0); j < (k); ++j) {
        rst.back().second.push_back({spe[j][i], j});
      }
      sort(rst.back().second.begin(), rst.back().second.end());
    }
  }
  sort(
      rst.begin(), rst.end(),
      [&](pair<int, vector<pair<int, int>>> a,
          pair<int, vector<pair<int, int>>> b) { return a.second < b.second; });
  for (int i = (0); i < (rst.size()); ++i) {
    int id = rst[i].first;
    vector<pair<int, int>> stuff = rst[i].second;
    int close = val[stuff[0].second];
    pair<int, int> tohang = {0, close};
    for (int j = (1); j < (stuff.size()); ++j) {
      int far = val[stuff[j].second];
      int cc = spe[stuff[0].second][far];
      int aa = stuff[0].first;
      int bb = stuff[j].first;
      if (aa + cc == bb) continue;
      int offtree = aa + bb - cc;
      if (offtree & 1) killthyself();
      offtree >>= 1;
      tohang = max(tohang, {aa - offtree, nthonpath(close, far, aa - offtree)});
    }
    tohang.first = stuff[0].first - tohang.first - 1;
    if (hang[tohang.second].size() > tohang.first + 1) {
      upd(id, hang[tohang.second][tohang.first]);
    } else {
      hang[tohang.second].push_back(id);
      if (hang[tohang.second].size() > tohang.first + 1) {
        upd(id, hang[tohang.second][tohang.first]);
      } else {
        killthyself();
      }
    }
  }
  for (int i = (0); i < (k); ++i) {
    for (int j = (0); j < (n); ++j) {
      if (dist(val[i], j) != spe[i][j]) killthyself();
    }
  }
  for (auto p : res) {
    cout << (p.first + 1) << " " << (p.second + 1) << "\n";
  }
  return 0;
}