Based on a peculiar incident at basketball practice, Akari came up with the following competitive programming problem!

You are given n points on the plane, no three of which are collinear. The i-th point initially has a label a_i, in such a way that the labels a_1, a_2, ..., a_n form a permutation of 1, 2, ..., n.

You are allowed to modify the labels through the following operation:

  1. Choose two distinct integers i and j between 1 and n. 
  2. Swap the labels of points i and j, and finally 
  3. Draw the segment between points i and j. 



A sequence of operations is valid if after applying all of the operations in the sequence in order, the k-th point ends up having the label k for all k between 1 and n inclusive, and the drawn segments don't intersect each other internally. Formally, if two of the segments intersect, then they must do so at a common endpoint of both segments.

In particular, all drawn segments must be distinct.

Find any valid sequence of operations, or say that none exist. 

Input

The first line contains an integer n (3 ≤ n ≤ 2000) — the number of points.

The i-th of the following n lines contains three integers x_i, y_i, a_i (-10^6 ≤ x_i, y_i ≤ 10^6, 1 ≤ a_i ≤ n), representing that the i-th point has coordinates (x_i, y_i) and initially has label a_i.

It is guaranteed that all points are distinct, no three points are collinear, and the labels a_1, a_2, ..., a_n form a permutation of 1, 2, ..., n.

Output

If it is impossible to perform a valid sequence of operations, print -1.

Otherwise, print an integer k (0 ≤ k ≤ (n(n - 1))/(2)) — the number of operations to perform, followed by k lines, each containing two integers i and j (1 ≤ i, j ≤ n, i≠ j) — the indices of the points chosen for the operation.

Note that you are not required to minimize or maximize the value of k.

If there are multiple possible answers, you may print any of them.

Examples

Input


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


Output


5
1 2
5 3
4 5
1 5
1 3


Input


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


Output


0

Note

The following animation showcases the first sample test case. The black numbers represent the indices of the points, while the boxed orange numbers represent their labels.

<image>

In the second test case, all labels are already in their correct positions, so no operations are necessary.
#include<bits/stdc++.h>
using namespace std ;

#define ll              long long 
#define pb              push_back
#define all(v)          v.begin(),v.end()
#define sz(a)           (ll)a.size()
#define F               first
#define S               second
#define INF             2000000000000000000
#define popcount(x)     __builtin_popcountll(x)
#define pll             pair<ll,ll> 
#define pii             pair<int,int>
#define ld              long double

const int M = 1000000007;
const int MM = 998244353;
const long double PI = acos(-1);

template<typename T, typename U> static inline void amin(T &x, U y){ if(y<x) x=y; }
template<typename T, typename U> static inline void amax(T &x, U y){ if(x<y) x=y; }
template<typename T, typename U> ostream& operator<<(ostream &os, const pair<T, U> &p)
{ 
    return os<<'('<<p.F<< ","<<p.S<<')'; 
}

template<typename T = long long>
struct Point
{
    T x,y;
    int id;
    Point() : x(0),y(0) {}
    Point(T x_,T y_): x(x_),y(y_) {}
    friend Point operator-(const Point &a, const Point &b) { return Point(a.x-b.x,a.y-b.y); }
    friend Point operator+(const Point &a, const Point &b) { return Point(a.x+b.x,a.y+b.y); }
    friend ostream& operator << (ostream &stream, const Point &P) { return stream<<"("<<P.x<<","<<P.y<<")"; }
    long long abs() { return x*x + y*y; }
};
template<typename T>
T dot(const Point<T> &a,const Point<T> &b)
{
    return a.x*b.x+a.y*b.y;
}
template<typename T>
T cross(const Point<T> &a,const Point<T> &b)
{
    return a.x*b.y-b.x*a.y;
}
using Pt = Point<ll> ;
struct AngCmp
{
    // centred at origin and anticlockwise relative to V;
    Pt origin,V;
    AngCmp(Pt a,Pt b): origin(a),V(b) {}

    bool operator() (const Pt &a,const Pt &b) const
    {
        Pt A = a-origin;
        Pt B = b-origin;

        bool x = cross(V,A)>=0;
        bool y = cross(V,B)>=0;
        if(x^y)
            return x;
        ll u = cross(A,B);
        if(u != 0)
            return u > 0;
        ll ux = dot(A,B);
        ll vx = dot(A,V);
        if(ux < 0)
            return vx > 0;
        // change here for same polar angle
        return A.abs()>B.abs();
    }
};

// AngCmp cmp(Pt(0,0),Pt(1,0)) <-- Like this

const int N = 2005;
int par[N],sz[N];
int parent(int i)
{
    return par[i] == i?i:parent(par[i]);
}
int comp;
bool merge(int x,int y)
{
    x = parent(x),y = parent(y);
    if(x == y)
        return false;
    if(sz[x] > sz[y])
        swap(x,y);
    par[x] = y;
    sz[y] += sz[x];
    --comp;
    return true;
}   

int _runtimeTerror_()
{
    int n;
    cin>>n;
    vector<int> p(n+1);
    comp = n;
    for(int i=1;i<=n;++i)
        par[i] = i,sz[i] = 1;
    vector<Pt> pts;
    for(int i=1;i<=n;++i)
    {
        Pt x;
        cin>>x.x>>x.y>>p[i];
        x.id = i;
        merge(i,p[i]);
        if(i == p[i])
            continue;
        pts.push_back(x);
    }
    if(pts.empty())
    {
        cout<<"0\n";return 0;
    }
    sort(all(pts),[&](Pt &x,Pt &y){
        return x.x < y.x;
    });
    Pt a = pts.back();
    pts.pop_back();
    AngCmp cmp(a,Pt(1,0));
    sort(all(pts),cmp);
    auto make = [&](int i,int j)
    {
        swap(p[i],p[j]);
    };
    vector<pii> ans;
    for(int i=0;i<sz(pts);++i)
    {
        if(merge(pts[i].id,pts[(i+1)%sz(pts)].id))
        {
            ans.push_back({pts[i].id,pts[(i+1)%sz(pts)].id});
            make(pts[i].id,pts[(i+1)%sz(pts)].id);
        }
    }
    int cur = a.id;
    while(p[cur] != cur)
    {
        ans.push_back({cur,p[cur]});
        make(cur,p[cur]);
    }
    // assert(comp == 1);
    cout<<sz(ans)<<"\n";
    for(auto &[j,k]:ans)
        cout<<j<<" "<<k<<"\n";
    return 0;
}

int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    #ifdef runSieve
        sieve();
    #endif
    #ifdef NCR
        initialize();
    #endif
    int TESTS=1;
    //cin>>TESTS;
    while(TESTS--)
        _runtimeTerror_();
    return 0;
}