/*
Condition:
- Sum of arrays must be divisible by k
- Maximum number must be smaller than or equal to (sum / k)
Solution:
- Divide array into k blocks
- ....
*/
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define FORD(i, a, b) for (int i = (a); i >= (b); --i)
#define REP(i, a) for (int i = 0; i < (a); ++i)
#define DEBUG(x) { cerr << #x << '=' << x << endl; }
#define Arr(a, l, r) { cerr << #a << " = {"; FOR(_, l, r) cerr << ' ' << a[_]; cerr << "}\n"; }
#define N 1001000
#define pp pair<long long, int>
#define endl '\n'
#define IO ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define taskname ""
#define bit(S, i) (((S) >> (i)) & 1)
using namespace std;
int n, k;
vector< vector<pp> > block;
vector< vector<long long> > res;
long long a[N];
int main() {
#ifdef NERO
freopen("test.inp","r",stdin);
freopen("test.out","w",stdout);
double stime = clock();
#else
//freopen(taskname".inp","r",stdin);
//freopen(taskname".out","w",stdout);
#endif //NERO
IO;
cin >> n >> k;
long long sum = 0, mx = 0;
FOR(i, 1, n) {
cin >> a[i];
mx = max(mx, a[i]);
sum += a[i];
}
long long average = (sum / k);
if (sum % k || mx > average) {
cout << -1;
return 0;
}
vector<pp> now;
long long temp = 0;
FOR(i, 1, n) {
temp += a[i];
if (temp >= average) {
now.push_back(pp(a[i] - (temp - average), i));
block.push_back(now);
temp -= average;
now.clear();
if (temp) now.push_back(pp(temp, i));
} else now.push_back(pp(a[i], i));
}
while (sum > 0) {
vector<long long> ans;
long long subtract = 1e18;
REP(i, k) subtract = min(subtract, block[i].back().first);
sum -= subtract * k;
ans.push_back(subtract);
REP(i, k) {
block[i].back().first -= subtract;
ans.push_back(block[i].back().second);
if (block[i].back().first == 0) block[i].pop_back();
}
res.push_back(ans);
}
cout << (int) res.size() << '\n';
REP(i, res.size()) {
for (long long x : res[i]) cout << x << ' '; cout << '\n';
}
#ifdef NERO
double etime = clock();
cerr << "Execution time: " << (etime - stime) / CLOCKS_PER_SEC * 1000 << " ms.\n";
#endif // NERO
}
