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;
struct edge {
  int to, next;
  edge(int _to = 0, int _next = 0) : to(_to), next(_next) {}
} e[30010 << 1];
int n, m;
int g[30010], nume;
int f[30010];
int dis[210][30010], s[30010], l[30010];
vector<pair<int, int> > S[30010];
void addEdge(int u, int v) {
  e[nume] = edge(v, g[u]);
  g[u] = nume++;
}
namespace Tree {
int size[30010], son[30010], top[30010], pre[30010], dep[30010], w[30010],
    pt[30010], numw;
void dfs(int x, int p) {
  size[x] = 1;
  son[x] = 0;
  pre[x] = p;
  for (int i = g[x]; ~i; i = e[i].next)
    if (e[i].to ^ p) {
      dep[e[i].to] = dep[x] + 1;
      dfs(e[i].to, x);
      size[x] += size[e[i].to];
      if (size[e[i].to] > size[son[x]]) son[x] = e[i].to;
    }
}
void dfs2(int x, int p) {
  pt[w[x] = ++numw] = x;
  if (son[x]) {
    top[son[x]] = top[x];
    dfs2(son[x], x);
  }
  for (int i = g[x]; ~i; i = e[i].next)
    if (e[i].to != p && e[i].to != son[x]) {
      top[e[i].to] = e[i].to;
      dfs2(e[i].to, x);
    }
}
void build() {
  dfs(s[1], 0);
  top[s[1]] = s[1];
  dfs2(s[1], 0);
}
int getLCA(int x, int y) {
  while (top[x] ^ top[y]) {
    if (dep[top[x]] < dep[top[y]]) x ^= y ^= x ^= y;
    x = pre[top[x]];
  }
  return dep[x] < dep[y] ? x : y;
}
int getDis(int x, int y) { return dep[x] + dep[y] - 2 * dep[getLCA(x, y)]; }
int getPos(int x, int p) {
  int s1 = s[x], s2 = s[x + 1];
  int len = l[x];
  if ((dis[x][p] + dis[x + 1][p] - len) & 1) return -1;
  int d1 = dis[x][p] - (dis[x][p] + dis[x + 1][p] - len) / 2;
  if (d1 < 0 || d1 > len) return -1;
  return d1;
}
int goUp(int x, int d) {
  while (dep[x] - dep[top[x]] + 1 <= d) {
    d -= dep[x] - dep[top[x]] + 1;
    x = pre[top[x]];
  }
  return pt[w[x] - d];
}
int go(int x, int y, int d) {
  int lca = getLCA(x, y);
  if (dep[x] - dep[lca] >= d) return goUp(x, d);
  d = dep[x] + dep[y] - 2 * dep[lca] - d;
  return goUp(y, d);
}
bool insert(int x) {
  if (m == 1) {
    S[s[1]].push_back(make_pair(dis[1][x], x));
    return 1;
  }
  int mindep = n;
  for (int i = 1; i < m; i++) {
    int len = l[i], pos = getPos(i, x);
    if (pos == -1) return 0;
    if (dis[i][x] - pos < mindep) mindep = dis[i][x] - pos;
  }
  for (int i = 1; i < m; i++) {
    int len = l[i], pos = getPos(i, x);
    if (dis[i][x] - pos == mindep) {
      int temp = go(s[i], s[i + 1], pos);
      S[temp].push_back(make_pair(dis[i][x] - pos, x));
      return 1;
    }
  }
  return 0;
}
bool gao_insert(int x) {
  int last = x, lastdep = 0;
  sort(S[x].begin(), S[x].end());
  for (int i = 0; i < S[x].size(); i++) {
    if (S[x][i].first == lastdep) return 0;
    if (S[x][i].first - lastdep > 1) {
      if (i == 0) return 0;
      last = S[x][i - 1].second;
      lastdep = S[x][i - 1].first;
      if (S[x][i].first - lastdep > 1) return 0;
    }
    addEdge(last, S[x][i].second);
    addEdge(S[x][i].second, last);
  }
  return 1;
}
}  // namespace Tree
namespace Check {
int d[30010];
bool check(int id) {
  int S = s[id];
  static queue<int> Q;
  for (int i = 1; i <= n; i++) d[i] = -1;
  Q.push(S);
  d[S] = 0;
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = g[x]; ~i; i = e[i].next)
      if (d[e[i].to] == -1) {
        d[e[i].to] = d[x] + 1;
        Q.push(e[i].to);
      }
  }
  for (int i = 1; i <= n; i++)
    if (d[i] != dis[id][i]) return 0;
  return 1;
}
}  // namespace Check
int getf(int x) {
  if (f[x] == x) return x;
  return f[x] = getf(f[x]);
}
bool gao(int x) {
  static int pt[30010];
  int s1 = 0, s2 = 0;
  for (int i = 1; i <= n; i++) {
    if (dis[x][i] == 0) s1 = i;
    if (dis[x + 1][i] == 0) s2 = i;
  }
  if (!s1 || !s2) return 0;
  int len = dis[x][s2];
  l[x] = len;
  for (int i = 0; i <= len; i++) pt[i] = 0;
  for (int i = 1; i <= n; i++)
    if (dis[x][i] + dis[x + 1][i] == len) {
      if (pt[dis[x][i]]) return 0;
      pt[dis[x][i]] = i;
    }
  for (int i = 1; i <= len; i++) {
    if (!pt[i - 1] || !pt[i]) return 0;
    if (getf(pt[i - 1]) != getf(pt[i])) {
      f[f[pt[i - 1]]] = f[pt[i]];
      addEdge(pt[i - 1], pt[i]);
      addEdge(pt[i], pt[i - 1]);
    }
  }
  return 1;
}
int main() {
  memset(g, -1, sizeof g);
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) f[i] = i;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= n; j++) {
      scanf("%d", &dis[i][j]);
      if (!dis[i][j]) s[i] = j;
    }
  for (int i = 1; i < m; i++)
    if (!gao(i)) {
      puts("-1");
      return 0;
    }
  Tree::build();
  for (int i = 1; i <= n; i++)
    if (getf(i) != getf(s[1]))
      if (!Tree::insert(i)) {
        puts("-1");
        return 0;
      }
  for (int i = 1; i <= n; i++)
    if (!Tree::gao_insert(i)) {
      puts("-1");
      return 0;
    }
  for (int i = 1; i <= m; i++)
    if (!Check::check(i)) {
      puts("-1");
      return 0;
    }
  Tree::build();
  for (int i = 1; i <= n; i++)
    if (i != s[1]) printf("%d %d\n", Tree::pre[i], i);
  return 0;
}