#include <bits/stdc++.h>
using namespace std;
const int MAXN = 25e2 + 2;
int n, c[MAXN][MAXN], par[MAXN], sz;
vector<pair<int, int>> adj[MAXN];
pair<int, int> sec[MAXN * MAXN];
bool cmp(pair<int, int> u, pair<int, int> v) {
  return c[u.first][u.second] < c[v.first][v.second];
}
int get_par(int v) { return par[v] == v ? v : par[v] = get_par(par[v]); }
bool dfs(int v, int ss, int p = -1, int mn = 0) {
  if (v ^ ss && c[v][ss] > mn) return true;
  for (auto u : adj[v])
    if (u.first ^ p && dfs(u.first, ss, v, max(mn, u.second))) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) scanf("%d", &c[i][j]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (c[i][j] ^ c[j][i] || (i == j && c[i][j]))
        return printf("NOT MAGIC\n"), 0;
  ;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      if (i ^ j) sec[sz++] = {i, j};
  iota(par, par + n, 0);
  sort(sec, sec + sz, cmp);
  for (int i = 0; i < sz; i++)
    if (get_par(sec[i].second) != get_par(sec[i].first)) {
      adj[sec[i].second].push_back(
          {sec[i].first, c[sec[i].second][sec[i].first]}),
          adj[sec[i].first].push_back(
              {sec[i].second, c[sec[i].first][sec[i].second]});
      par[get_par(sec[i].first)] = get_par(sec[i].second);
    }
  for (int i = 0; i < n; i++)
    if (dfs(i, i)) return printf("NOT MAGIC\n"), 0;
  ;
  printf("MAGIC\n");
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

You're given a matrix A of size n × n.

Let's call the matrix with nonnegative elements magic if it is symmetric (so aij = aji), aii = 0 and aij ≤ max(aik, ajk) for all triples i, j, k. Note that i, j, k do not need to be distinct.

Determine if the matrix is magic.

As the input/output can reach very huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains integer n (1 ≤ n ≤ 2500) — the size of the matrix A.

Each of the next n lines contains n integers aij (0 ≤ aij < 109) — the elements of the matrix A.

Note that the given matrix not necessarily is symmetric and can be arbitrary.

Output

Print ''MAGIC" (without quotes) if the given matrix A is magic. Otherwise print ''NOT MAGIC".

Examples

Input

3
0 1 2
1 0 2
2 2 0


Output

MAGIC


Input

2
0 1
2 3


Output

NOT MAGIC


Input

4
0 1 2 3
1 0 3 4
2 3 0 5
3 4 5 0


Output

NOT MAGIC

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 25e2 + 2;
int n, c[MAXN][MAXN], par[MAXN], sz;
vector<pair<int, int>> adj[MAXN];
pair<int, int> sec[MAXN * MAXN];
bool cmp(pair<int, int> u, pair<int, int> v) {
  return c[u.first][u.second] < c[v.first][v.second];
}
int get_par(int v) { return par[v] == v ? v : par[v] = get_par(par[v]); }
bool dfs(int v, int ss, int p = -1, int mn = 0) {
  if (v ^ ss && c[v][ss] > mn) return true;
  for (auto u : adj[v])
    if (u.first ^ p && dfs(u.first, ss, v, max(mn, u.second))) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) scanf("%d", &c[i][j]);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (c[i][j] ^ c[j][i] || (i == j && c[i][j]))
        return printf("NOT MAGIC\n"), 0;
  ;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      if (i ^ j) sec[sz++] = {i, j};
  iota(par, par + n, 0);
  sort(sec, sec + sz, cmp);
  for (int i = 0; i < sz; i++)
    if (get_par(sec[i].second) != get_par(sec[i].first)) {
      adj[sec[i].second].push_back(
          {sec[i].first, c[sec[i].second][sec[i].first]}),
          adj[sec[i].first].push_back(
              {sec[i].second, c[sec[i].first][sec[i].second]});
      par[get_par(sec[i].first)] = get_par(sec[i].second);
    }
  for (int i = 0; i < n; i++)
    if (dfs(i, i)) return printf("NOT MAGIC\n"), 0;
  ;
  printf("MAGIC\n");
  return 0;
}

```