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;
int n, q;
const int BLOCK = 256;
const int maxN = (1 << 12) + 10;
int a[maxN];
int CUR;
vector<pair<int, int> > add;
int get(int x, int y) {
  add.emplace_back(x, y);
  return ++CUR;
}
vector<vector<int> > build_full(vector<int>& inds, int l, int r) {
  int N = inds.size();
  assert(N != 0);
  assert(inds.size() == r - l + 1);
  if (inds.size() == 1) {
    vector<vector<int> > d(N, vector<int>(N, -1));
    d[0][0] = inds[0];
    return d;
  }
  int mid = (l + r) / 2;
  vector<int> f1, f2;
  for (int& p : inds) {
    if (a[p] <= mid) {
      f1.emplace_back(p);
    } else {
      f2.emplace_back(p);
    }
  }
  auto d1 = build_full(f1, l, mid);
  auto d2 = build_full(f2, mid + 1, r);
  vector<vector<int> > d(N, vector<int>(N, -1));
  for (int x = 0; x < inds.size(); x++) {
    int pl_1 = 0;
    while (pl_1 < f1.size() && f1[pl_1] < inds[x]) pl_1++;
    int pr_1 = pl_1 - 1;
    int pl_2 = 0;
    while (pl_2 < f2.size() && f2[pl_2] < inds[x]) pl_2++;
    int pr_2 = pl_2 - 1;
    for (int y = x; y < inds.size(); y++) {
      while (pr_1 + 1 < f1.size() && f1[pr_1 + 1] <= inds[y]) pr_1++;
      while (pr_2 + 1 < f2.size() && f2[pr_2 + 1] <= inds[y]) pr_2++;
      if (pl_1 > pr_1) {
        assert(pl_2 <= pr_2);
        d[x][y] = d2[pl_2][pr_2];
      } else if (pl_2 > pr_2) {
        assert(pl_1 <= pr_1);
        d[x][y] = d1[pl_1][pr_1];
      } else {
        d[x][y] = get(d1[pl_1][pr_1], d2[pl_2][pr_2]);
      }
    }
  }
  return d;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  CUR = n;
  vector<vector<vector<int> > > vals;
  vector<vector<int> > inds;
  for (int l = 1; l <= n; l += BLOCK) {
    int r = min(l + BLOCK - 1, n);
    vector<int> gg;
    for (int p = 1; p <= n; p++) {
      if (a[p] >= l && a[p] <= r) gg.emplace_back(p);
    }
    inds.emplace_back(gg);
    assert(gg.size() == r - l + 1);
    vals.emplace_back(build_full(gg, l, r));
  }
  vector<int> ans;
  for (int i = 0; i < q; i++) {
    int from, to;
    cin >> from >> to;
    int T = -1;
    for (int z = 0; z < inds.size(); z++) {
      int from1 =
          lower_bound(inds[z].begin(), inds[z].end(), from) - inds[z].begin();
      int to1 =
          upper_bound(inds[z].begin(), inds[z].end(), to) - inds[z].begin() - 1;
      if (from1 > to1) continue;
      if (T == -1) {
        T = vals[z][from1][to1];
      } else {
        T = get(T, vals[z][from1][to1]);
      }
    }
    assert(T != -1);
    ans.emplace_back(T);
  }
  cout << CUR << '\n';
  for (auto& it : add) {
    cout << it.first << " " << it.second << '\n';
  }
  for (int v : ans) cout << v << " ";
  return 0;
}