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=100005;
using namespace std;
int main()
{
    int n,x;
    int vis[Max],p[Max];
    while(~scanf("%d",&n))
    {
        int k = 0,i,j;
        vector<int>G[Max];
        memset(vis,0,sizeof(vis));
        for (i = 0; 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] = 1;
                p[k++] = x;
            }

        }
        int cnt = 0;
        memset(vis,-1,sizeof(vis));
        for (j = 0; j < k; j++)
        {
            int len = G[p[j]].size();
            if (len==1)
            {
                ++cnt;
                vis[p[j]]= 0;
                continue;
            }
            int d = G[p[j]][1]-G[p[j]][0];//求公差
            for (i = 1; i < len; i++)
            {
                if (G[p[j]][i]-G[p[j]][i-1]!=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
",cnt);
        sort(p, p+k);
        for (i = 0; i < k ; i++)
        {
            if (vis[p[i]]!=-1)
            {
                printf("%d %d
",p[i],vis[p[i]]);
            }
        }
    }
    return 0;
}