One day Jeff got hold of an integer sequence a1, a2, ..., an 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 ai = x). These numbers, sorted in the increasing order, must form an arithmetic progression.
Help Jeff, find all x that meet the problem conditions.
The first line contains integer n (1 ≤ n ≤ 105). The next line contains integers a1, a2, ..., an(1 ≤ ai ≤ 105). The numbers are separated by spaces.
In the first line print integer t — the number of valid x. On each of the next t lines print two integers x and px, where x is current suitable value, px is the common difference between numbers in the progression (if x occurs exactly once in the sequence, px must equal 0). Print the pairs in the order of increasing x.
1
2
1
2 0
8
1 2 1 3 1 2 1 5
4
1 2
2 4
3 0
5 0
In the first test 2 occurs exactly once in the sequence, ergo p2 = 0.
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n;
int a[maxn];
struct node {
int st;// 端点
int dx;// 差值
int cnt;// 计数
int fg;
}td[maxn];
map<int, int>mp;
int main() {
//ios::sync_with_stdio(0);
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
td[a[i]].cnt++;
if (td[a[i]].cnt == 1) {
td[a[i]].st = i; td[a[i]].fg = 1;
}
else {
if (td[a[i]].cnt == 2) {
td[a[i]].dx = i - td[a[i]].st;
td[a[i]].st = i;
}
else {
if (td[a[i]].dx != i - td[a[i]].st) {
td[a[i]].fg = 0;
}
else {
td[a[i]].dx = i - td[a[i]].st;
td[a[i]].st = i;
}
}
}
}
sort(a + 1, a + 1 + n);
int tot = unique(a + 1, a + 1 + n) - a - 1;
bool fg = 0; int ct = 0;
for (int i = 1; i <= tot; i++) {
if (td[a[i]].fg == 1) {
ct++;
}
}
if (ct == 0) {
cout << 0 << endl; return 0;
}
cout << ct << endl;
for (int i = 1; i <= tot; i++) {
// cout << a[i] << ' ' << td[a[i]].fg << ' ' << td[a[i]].dx << endl;
if (td[a[i]].cnt == 1) {
cout << a[i] << ' ' << 0 << endl;
}
else if (td[a[i]].cnt > 1 && td[a[i]].fg == 1) {
cout << a[i] << ' ' << td[a[i]].dx << endl;
}
}
return 0;
}