#include<bits/stdc++.h>
using namespace std;
#define int long long
#define in array<int, 2>
#define pb push_back
#define pob pop_back
#define fast() ios_base::sync_with_stdio(false); cin.tie(NULL)
const int MX = 1e6+5;
const int INF = 1e17;
const int LMX = 3e6+4;
int a[MX];
in b[MX];
pair<int, vector<int>> op[LMX];
signed main()
{
fast();
int n, k; cin >> n >> k;
b[0] = {INF, 0};
int S = 0;
for(int i = 1; i <= n; i++)
{
cin >> b[i][0]; b[i][1] = i;
S+=b[i][0];
}
sort(b+1, b+n+1);
reverse(b+1, b+n+1);
for(int i = 1; i <= n; i++)
a[i] = b[i][0];
if((S%k) || (a[1] > (S/k)))
{
cout << "-1\n";
return 0;
}
int CUR = 0;
int g = n;
int s = 1;
while(true)
{
//Currently we want to achieve a[s] = S/k, previous ones are already S/k
//reduce S to a[i]*k
if(a[s] == (S/k))
{
s++;
continue;
}
if(a[s] == 0)
{
s--;
break;
}
while(a[g] == 0)
g--;
int D = min(a[g], (S/k)-a[s]);
for(int i = 1; i < s; i++)
{
op[CUR].second.pb(i);
a[i]-=D;
}
int d = g;
while(op[CUR].second.size() != k)
{
op[CUR].second.pb(d); a[d]-=D;
d--;
}
op[CUR].first = D;
S-=(D*k);
CUR++;
}
if(a[1])
{
for(int i = 1; i <= k; i++)
op[CUR].second.pb(i);
op[CUR].first = a[1];
CUR++;
}
cout << CUR << "\n";
for(int i = 0; i < CUR; i++)
{
cout << op[i].first << " ";
for(auto x: op[i].second)
cout << b[x][1] << " ";
cout << "\n";
}
return 0;
}
