B. Jeff and Periods(cf)

B. Jeff and Periods

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Jeff got hold of an integer sequence a₁, a₂, ..., a_n of length n. The boy immediately decided to analyze the sequence. For that, he needs to find all values of x, for which these conditions hold:

• x occurs in sequence a.
• Consider all positions of numbers x in the sequence a (such i, that a_i = x). These numbers, sorted in the increasing order, must form an arithmetic progression.

Help Jeff, find all x that meet the problem conditions.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵). The next line contains integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁵). The numbers are separated by spaces.

Output

In the first line print integer t — the number of valid x. On each of the next t lines print two integers x and p_x, where x is current suitable value, p_x is the common difference between numbers in the progression (if x occurs exactly once in the sequence, p_x must equal 0). Print the pairs in the order of increasing x.

Sample test(s)

Input

1
2

Output

1
2 0

Input

8
1 2 1 3 1 2 1 5

Output

4
1 2
2 4
3 0
5 0

 #include <stdio.h>
 #include <string.h>
 #include <vector>
 #include <algorithm>
 const int Max=;
 using namespace std;
 int main()
 {
     int n,x;
     int vis[Max],p[Max];
     while(~scanf("%d",&n))
     {
         int k = ,i,j;
         vector<int>G[Max];
         memset(vis,,sizeof(vis));
         for (i = ; i < n; i++)
         {
             scanf("%d",&x);
             G[x].push_back(i);//将所有x的位置存入vector中
             if (!vis[x])
             {
                 vis[x] = ;
                 p[k++] = x;
             }

         }
         int cnt = ;
         memset(vis,-,sizeof(vis));
         for (j = ; j < k; j++)
         {
             int len = G[p[j]].size();
             if (len==)
             {
                 ++cnt;
                 vis[p[j]]= ;
                 continue;
             }
             int d = G[p[j]][]-G[p[j]][];//求公差
             for (i = ; i < len; i++)
             {
                 if (G[p[j]][i]-G[p[j]][i-]!=d)
                     break;
             }
             if (i >= len)//说明p[j]的各位置是等差数列
             {
                 ++cnt;
                 vis[p[j]] = d;//表示p[j]各位置的公差为d
             }
         }
         printf("%d\n",cnt);
         sort(p, p+k);
         for (i = ; i < k ; i++)
         {
             if (vis[p[i]]!=-)
             {
                 printf("%d %d\n",p[i],vis[p[i]]);
             }
         }
     }
     return ;
 }