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 num;
int a[1 << 12], b[1 << 12], k[1 << 16];
vector<pair<int, int> > sol;
vector<int> tree[1 << 13];
map<pair<int, int>, int> M[1 << 13];
int build(int s, int e, int i) {
  if (s == e) {
    tree[i].push_back(b[s]);
    M[i][make_pair(0, 0)] = b[s];
    return 0;
  }
  int mid = (s + e) / 2;
  build(s, mid, 2 * i + 1), build(mid + 1, e, 2 * i + 2);
  tree[i].resize(e - s + 1);
  merge(tree[2 * i + 1].begin(), tree[2 * i + 1].end(), tree[2 * i + 2].begin(),
        tree[2 * i + 2].end(), tree[i].begin());
  return 0;
}
int query(int s, int e, int l, int r, int i) {
  l = lower_bound(tree[i].begin(), tree[i].end(), l) - tree[i].begin();
  r = upper_bound(tree[i].begin(), tree[i].end(), r) - tree[i].begin() - 1;
  if (l > r)
    return -1;
  else if (M[i].count(make_pair(l, r)))
    return M[i][make_pair(l, r)];
  int mid = (s + e) / 2;
  int a = query(s, mid, tree[i][l], tree[i][r], 2 * i + 1),
      b = query(mid + 1, e, tree[i][l], tree[i][r], 2 * i + 2);
  if (a == -1)
    M[i][make_pair(l, r)] = b;
  else if (b == -1)
    M[i][make_pair(l, r)] = a;
  else {
    sol.push_back(make_pair(a, b));
    M[i][make_pair(l, r)] = num++;
  }
  return M[i][make_pair(l, r)];
}
int main() {
  int i;
  int n, q, l, r;
  scanf("%d %d", &n, &q), num = n;
  for (i = 0; i < n; i++) scanf("%d", &a[i]), b[--a[i]] = i;
  build(0, n - 1, 0);
  for (i = 0; i < q; i++) {
    scanf("%d %d", &l, &r);
    l--, r--;
    k[i] = query(0, n - 1, l, r, 0);
  }
  printf("%d\n", num);
  for (i = 0; i < sol.size(); i++)
    printf("%d %d\n", sol[i].first + 1, sol[i].second + 1);
  for (i = 0; i < q; i++) printf("%d ", k[i] + 1);
  printf("\n");
  return 0;
}