Problem: You are given a permutation a_1, a_2, ..., a_n of numbers from 1 to n. Also, you have n sets S_1,S_2,..., S_n, where S_i=\\{a_i\}. Lastly, you have a variable cnt, representing the current number of sets. Initially, cnt = n.

We define two kinds of functions on sets:

f(S)=min_{u∈ S} u;

g(S)=max_{u∈ S} u.

You can obtain a new set by merging two sets A and B, if they satisfy g(A)<f(B) (Notice that the old sets do not disappear).

Formally, you can perform the following sequence of operations:

  * cnt← cnt+1;

  * S_{cnt}=S_u∪ S_v, you are free to choose u and v for which 1≤ u, v < cnt and which satisfy g(S_u)<f(S_v).




You are required to obtain some specific sets.

There are q requirements, each of which contains two integers l_i,r_i, which means that there must exist a set S_{k_i} (k_i is the ID of the set, you should determine it) which equals \\{a_u∣ l_i≤ u≤ r_i\}, which is, the set consisting of all a_i with indices between l_i and r_i.

In the end you must ensure that cnt≤ 2.2× 10^6. Note that you don't have to minimize cnt. It is guaranteed that a solution under given constraints exists.

Input

The first line contains two integers n,q (1≤ n ≤ 2^{12},1 ≤ q ≤ 2^{16}) — the length of the permutation and the number of needed sets correspondently.

The next line consists of n integers a_1,a_2,⋅⋅⋅, a_n (1≤ a_i≤ n, a_i are pairwise distinct) — given permutation.

i-th of the next q lines contains two integers l_i,r_i (1≤ l_i≤ r_i≤ n), describing a requirement of the i-th set.

Output

It is guaranteed that a solution under given constraints exists.

The first line should contain one integer cnt_E (n≤ cnt_E≤ 2.2× 10^6), representing the number of sets after all operations.

cnt_E-n lines must follow, each line should contain two integers u, v (1≤ u, v≤ cnt', where cnt' is the value of cnt before this operation), meaning that you choose S_u, S_v and perform a merging operation. In an operation, g(S_u)<f(S_v) must be satisfied.

The last line should contain q integers k_1,k_2,⋅⋅⋅,k_q (1≤ k_i≤ cnt_E), representing that set S_{k_i} is the ith required set.

Please notice the large amount of output.

Examples

Input


3 2
1 3 2
2 3
1 3


Output


6
3 2
1 3
5 2
4 6 

Input


2 4
2 1
1 2
1 2
1 2
1 1


Output


5
2 1
2 1
2 1
5 3 3 1

Note

In the first sample:

We have S_1=\{1\},S_2=\{3\},S_3=\{2\} initially.

In the first operation, because g(S_3)=2<f(S_2)=3, we can merge S_3,S_2 into S_4=\{2,3\}.

In the second operation, because g(S_1)=1<f(S_3)=2, we can merge S_1,S_3 into S_5=\{1,2\}.

In the third operation, because g(S_5)=2<f(S_2)=3, we can merge S_5,S_2 into S_6=\{1,2,3\}.

For the first requirement, S_4=\{2,3\}=\\{a_2,a_3\}, satisfies it, thus k_1=4.

For the second requirement, S_6=\{1,2,3\}=\\{a_1,a_2,a_3\}, satisfies it, thus k_2=6

Notice that unused sets, identical sets, outputting the same set multiple times, and using sets that are present initially are all allowed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
int n, q, B, ind, a[N], aa[N];
vector<vector<int> > f[N];
int nid;
vector<int> pos[260], res;
vector<pair<int, int> > ans;
int mer(int k1, int k2) {
  if (!k1 || !k2) return k1 + k2;
  ans.push_back(make_pair(k1, k2));
  return ++ind;
}
vector<vector<int> > bud(int l, int r, vector<int> npos) {
  if (l == r) {
    return {{npos[0]}};
  }
  int mid = (l + r) >> 1;
  vector<int> posl, posr;
  for (auto x : npos) (a[x] <= mid ? posl : posr).push_back(x);
  auto resl = bud(l, mid, posl), resr = bud(mid + 1, r, posr),
       res = vector<vector<int> >(r - l + 1, vector<int>(r - l + 1));
  for (int i = (0); i <= (((int)(npos).size()) - 1); ++i)
    for (int j = (i); j <= (((int)(npos).size()) - 1); ++j) {
      int L = lower_bound((posl).begin(), (posl).end(), npos[i]) - posl.begin();
      int R =
          upper_bound((posl).begin(), (posl).end(), npos[j]) - posl.begin() - 1;
      int& t = res[i][j];
      t = 0;
      if (L <= R) {
        t = resl[L][R];
      }
      L = lower_bound((posr).begin(), (posr).end(), npos[i]) - posr.begin();
      R = upper_bound((posr).begin(), (posr).end(), npos[j]) - posr.begin() - 1;
      if (L <= R) {
        t = mer(t, resr[L][R]);
      }
    }
  return res;
}
void sol(int L, int R) {
  pos[nid].clear();
  for (int i = (L); i <= (R); ++i) pos[nid].push_back(aa[i]);
  sort((pos[nid]).begin(), (pos[nid]).end());
  f[nid] = bud(L, R, pos[nid]);
}
int main() {
  scanf("%d%d", &n, &q);
  B = 1 << 8;
  for (int i = (1); i <= (n); ++i) scanf("%d", a + i), aa[a[i]] = i;
  nid = 1, ind = n;
  for (int i = 1; i <= n; i += B) {
    sol(i, min(n, i + B - 1));
    ++nid;
  }
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    nid = 0;
    int now = 0;
    for (int i = 1; i <= n; i += B) {
      ++nid;
      auto L = lower_bound((pos[nid]).begin(), (pos[nid]).end(), l) -
               pos[nid].begin();
      auto R = upper_bound((pos[nid]).begin(), (pos[nid]).end(), r) -
               pos[nid].begin() - 1;
      if (L <= R) {
        now = mer(now, f[nid][L][R]);
      }
    }
    res.push_back(now);
  }
  printf("%d\n", ((int)(ans).size()) + n);
  for (auto x : ans) printf("%d %d\n", x.first, x.second);
  for (auto x : res) printf("%d ", x);
  return 0;
}