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 B = 256;
int n, q, tree[555][266][266], val[22][266][266], pos[4444], a[4444], cnt, cpt,
    cur;
int arr[22][266], len[22], l, r, res[222222], ll, rr;
int ar[666], br[666], cr[666], la, lb, lc;
pair<int, int> ans[2222222];
char prt[55555555];
void merge(int l, int mid, int r, int node) {
  la = lb = lc = 0;
  for (int i = l; i <= mid; i++) ar[++la] = pos[i];
  for (int i = mid + 1; i <= r; i++) br[++lb] = pos[i];
  for (int i = l; i <= r; i++) cr[++lc] = pos[i];
  sort(ar + 1, ar + la + 1);
  sort(br + 1, br + lb + 1);
  sort(cr + 1, cr + lc + 1);
  for (int i = 1; i <= lc; i++) {
    for (int j = i; j <= lc; j++) {
      l = lower_bound(ar + 1, ar + la + 1, cr[i]) - ar;
      r = lower_bound(ar + 1, ar + la + 1, cr[j] + 1) - ar - 1;
      ll = lower_bound(br + 1, br + lb + 1, cr[i]) - br;
      rr = lower_bound(br + 1, br + lb + 1, cr[j] + 1) - br - 1;
      if (l > r)
        tree[node][i][j] = tree[(node << 1) | 1][ll][rr];
      else if (ll > rr)
        tree[node][i][j] = tree[node << 1][l][r];
      else {
        ++cnt;
        ans[cnt] =
            make_pair(tree[node << 1][l][r], tree[(node << 1) | 1][ll][rr]);
        tree[node][i][j] = cnt;
      }
    }
  }
}
void solve(int l, int r, int node) {
  if (l == r) {
    tree[node][1][1] = pos[l];
    return;
  }
  int mid = (l + r) >> 1;
  solve(l, mid, node << 1);
  solve(mid + 1, r, (node << 1) | 1);
  merge(l, mid, r, node);
}
void print(int x) {
  if (!x) {
    prt[cpt++] = '0';
    return;
  }
  while (x) {
    prt[cpt++] = '0' + (x % 10);
    x /= 10;
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  cnt = n;
  for (int i = 0; i * B + 1 <= n; i++) {
    for (int j = i * B + 1; j <= min(n, (i + 1) * B); j++) {
      arr[i][++len[i]] = pos[j];
    }
    sort(arr[i] + 1, arr[i] + len[i] + 1);
    solve(i * B + 1, min(n, (i + 1) * B), 1);
    for (int j = 1; j <= len[i]; j++) {
      for (int k = j; k <= len[i]; k++) val[i][j][k] = tree[1][j][k];
    }
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &l, &r);
    cur = -1;
    for (int j = 0; j <= (n - 1) / B; j++) {
      ll = lower_bound(arr[j] + 1, arr[j] + len[j] + 1, l) - arr[j];
      rr = lower_bound(arr[j] + 1, arr[j] + len[j] + 1, r + 1) - arr[j] - 1;
      if (ll > rr) continue;
      if (cur == -1)
        cur = val[j][ll][rr];
      else {
        ++cnt;
        ans[cnt] = make_pair(cur, val[j][ll][rr]);
        cur = cnt;
      }
    }
    res[i] = cur;
  }
  cpt = 1;
  prt[0] = '\n';
  for (int i = q; i >= 1; i--) {
    print(res[i]);
    if (i > 1) prt[cpt++] = ' ';
  }
  prt[cpt++] = '\n';
  for (int i = cnt; i > n; i--) {
    print(ans[i].second);
    prt[cpt++] = ' ';
    print(ans[i].first);
    prt[cpt++] = '\n';
  }
  print(cnt);
  reverse(prt, prt + cpt);
  fwrite(prt, 1, cpt, stdout);
  return 0;
}