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 = (1 << 12) + 7;
const int Q = (1 << 16) + 7;
const int S = sqrt(Q);
int total;
vector<pair<int, int>> merged;
int merge(int v, int u) {
  if (v == -1 || u == -1) {
    return v + u + 1;
  }
  merged.push_back({v, u});
  return total++;
}
vector<int> b;
vector<vector<int>> rec(vector<int>& block, int l, int r) {
  int n = block.size();
  vector<vector<int>> res(n + 1, vector<int>(n + 1, -1));
  if (!n) {
    return res;
  }
  if (r - l == 1) {
    assert(n == 1);
    res[0][1] = block[0];
    return res;
  }
  int mid = (l + r) / 2;
  vector<int> lblock, rblock;
  for (int first : block) {
    if (b[first] < mid) {
      lblock.push_back(first);
    } else {
      rblock.push_back(first);
    }
  }
  auto lseg = rec(lblock, l, mid);
  auto rseg = rec(rblock, mid, r);
  int l_left = 0, l_right = 0;
  for (int i = 0; i < n; ++i) {
    int r_left = l_left, r_right = l_right;
    for (int j = i; j < n; ++j) {
      if (b[block[j]] < mid) {
        ++r_left;
      } else {
        ++r_right;
      }
      res[i][j + 1] = merge(lseg[l_left][r_left], rseg[l_right][r_right]);
    }
    if (b[block[i]] < mid) {
      ++l_left;
    } else {
      ++l_right;
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  total = n;
  vector<pair<int, int>> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a.begin(), a.end());
  b.resize(n);
  for (int i = 0; i < n; ++i) {
    b[a[i].second] = i;
  }
  vector<vector<int>> blocks;
  vector<vector<vector<int>>> idxs;
  vector<int> pos(n, -1);
  for (int i = 0; i < n; i += S) {
    vector<int> block;
    for (int j = 0; j < n; ++j) {
      if (i <= b[j] && b[j] < i + S) {
        block.push_back(j);
      }
    }
    blocks.push_back(block);
    idxs.push_back(rec(block, i, i + S));
  }
  vector<int> res(q, -1);
  for (int it = 0; it < q; ++it) {
    int l, r;
    cin >> l >> r;
    --l, --r;
    int now = -1;
    int cur = 0;
    for (int i = 0; i < n; i += S) {
      int lv = lower_bound(blocks[cur].begin(), blocks[cur].end(), l) -
               blocks[cur].begin();
      int rv = upper_bound(blocks[cur].begin(), blocks[cur].end(), r) -
               blocks[cur].begin();
      now = merge(now, idxs[cur][lv][rv]);
      ++cur;
    }
    res[it] = now;
  }
  cout << total << endl;
  for (auto t : merged) {
    cout << t.first + 1 << " " << t.second + 1 << "\n";
  }
  for (int i = 0; i < q; ++i) {
    cout << res[i] + 1 << " ";
  }
  cout << "\n";
}