CF352B Jeff and Periods 模拟

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.

Examples

Input

Copy

1
2

Output

Copy

1
2 0

Input

Copy

8
1 2 1 3 1 2 1 5

Output

Copy

4
1 2
2 4
3 0
5 0

Note

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;
}