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;
int a[4097];
int topos[4097];
int cnt;
vector<pair<int, int> > ans;
int anspos[100001];
struct seg {
  struct node {
    int l, r;
    vector<vector<int> > topos;
    vector<int> rg;
  } seg[200001];
  int bins(int p, int val, int type) {
    int lo = 1, hi = seg[p].rg.size() + 1, m;
    while (lo < hi) {
      m = (lo + hi) / 2;
      if ((type == 1 && seg[p].rg[m - 1] <= val) ||
          (type == 2 && seg[p].rg[m - 1] < val))
        lo = m + 1;
      else
        hi = m;
    }
    if (type == 1) return lo - 1;
    return lo;
  }
  int query(int p, int l, int r) {
    if (seg[p].l + 1 == seg[p].r) {
      int x = bins(p, l, 2), y = bins(p, r, 1);
      if (y < x) return 0;
      seg[p].topos[x][y] = topos[seg[p].l];
      return seg[p].topos[x][y];
    } else {
      int x = bins(p, l, 2), y = bins(p, r, 1);
      if (y < x) return 0;
      if (seg[p].topos[x][y] == 0) {
        int retx = query(2 * p, l, r), rety = query(2 * p + 1, l, r);
        if (retx == 0)
          seg[p].topos[x][y] = rety;
        else if (rety == 0)
          seg[p].topos[x][y] = retx;
        else {
          cnt++;
          ans.push_back({retx, rety});
          seg[p].topos[x][y] = cnt;
        }
      }
      return seg[p].topos[x][y];
    }
  }
  void build(int p, int l, int r) {
    int i, j;
    seg[p].l = l, seg[p].r = r;
    for (i = 0; i <= r - l; i++) {
      vector<int> cur;
      for (j = 0; j <= r - l; j++) cur.push_back(0);
      seg[p].topos.push_back(cur);
    }
    for (i = l; i < r; i++) seg[p].rg.push_back(topos[i]);
    sort(seg[p].rg.begin(), seg[p].rg.end());
    if (l + 1 < r) {
      build(2 * p, l, (l + r) / 2);
      build(2 * p + 1, (l + r) / 2, r);
    }
  }
} ds;
int main() {
  int i;
  scanf("%d%d", &n, &q), cnt = n;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), topos[a[i]] = i;
  ds.build(1, 1, n + 1);
  for (i = 1; i <= q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    anspos[i] = ds.query(1, x, y);
  }
  printf("%d\n", cnt);
  for (i = 0; i < ans.size(); i++)
    printf("%d %d\n", ans[i].first, ans[i].second);
  for (i = 1; i < q; i++) printf("%d ", anspos[i]);
  printf("%d\n", anspos[q]);
  return 0;
}