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;
std::mt19937 rnd(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 1 << 12;
const int MAXQ = 1 << 16;
const int MAXCNT = 2200000;
const int MAXB = MAXN;
int n, nq;
int a[MAXN];
int ql[MAXQ], qr[MAXQ], qans[MAXQ];
pair<int, int> how[MAXCNT];
int cnt;
int pos[MAXN];
int bsz, nb;
int bl[MAXB], br[MAXB];
int bfst[MAXB][MAXN], blst[MAXB][MAXN];
int id[MAXN];
vector<vector<int>> bseg[MAXB];
int merge(int a, int b) {
  assert(cnt < MAXCNT);
  int ret = cnt++;
  how[ret] = make_pair(a, b);
  return ret;
}
vector<vector<int>> rec(int b, int l, int r, const vector<int> &who) {
  int sz = ((int)(who).size());
  vector<vector<int>> ret(sz, vector<int>(sz, -1));
  if (l == r) {
    assert(sz == 1 && a[who[0]] == l);
    ret[0][0] = who[0];
  } else {
    int m = l + (r - l) / 2;
    vector<int> lidx(sz, -1), ridx(sz, -1), lwho, rwho;
    for (int i = (0); i < (sz); ++i) {
      int id = who[i];
      if (a[id] <= m)
        lidx[i] = ((int)(lwho).size()), lwho.push_back(id);
      else
        ridx[i] = ((int)(rwho).size()), rwho.push_back(id);
    }
    vector<vector<int>> lres = rec(b, l, m, lwho);
    vector<vector<int>> rres = rec(b, m + 1, r, rwho);
    for (int i = (0); i < (sz); ++i) {
      int lmn = ((int)(lwho).size()), lmx = -1, rmn = ((int)(rwho).size()),
          rmx = -1;
      for (int j = (i); j < (sz); ++j) {
        int id = who[j];
        if (a[id] <= m)
          lmn = min(lmn, lidx[j]), lmx = max(lmx, lidx[j]);
        else
          rmn = min(rmn, ridx[j]), rmx = max(rmx, ridx[j]);
        int a = lmn <= lmx ? lres[lmn][lmx] : -1,
            b = rmn <= rmx ? rres[rmn][rmx] : -1;
        if (a != -1 && b != -1)
          ret[i][j] = merge(a, b);
        else if (a != -1)
          ret[i][j] = a;
        else if (b != -1)
          ret[i][j] = b;
        else
          assert(false);
      }
    }
  }
  return ret;
}
void solve() {
  cnt = n;
  for (int i = (0); i < (n); ++i) pos[a[i]] = i;
  bsz = min(n, (int)sqrt(2.0 * nq));
  nb = (n + bsz - 1) / bsz;
  for (int b = (0); b < (nb); ++b)
    bl[b] = b * bsz, br[b] = min((b + 1) * bsz, n) - 1;
  for (int b = (0); b < (nb); ++b)
    for (int i = n - 1; i >= 0; --i)
      bfst[b][i] = bl[b] <= a[i] && a[i] <= br[b] ? i
                   : i + 1 < n                    ? bfst[b][i + 1]
                                                  : n;
  for (int b = (0); b < (nb); ++b)
    for (int i = (0); i < (n); ++i)
      blst[b][i] = bl[b] <= a[i] && a[i] <= br[b] ? i
                   : i - 1 >= 0                   ? blst[b][i - 1]
                                                  : -1;
  for (int b = (0); b < (nb); ++b) {
    vector<int> who;
    for (int i = (0); i < (n); ++i)
      if (bl[b] <= a[i] && a[i] <= br[b]) who.push_back(i);
    for (int i = (0); i < (((int)(who).size())); ++i) id[who[i]] = i;
    bseg[b] = rec(b, bl[b], br[b], who);
  }
  for (int i = (0); i < (nq); ++i) {
    vector<int> parts;
    for (int b = (0); b < (nb); ++b) {
      int l = bfst[b][ql[i]];
      int r = blst[b][qr[i]];
      if (l > r) continue;
      parts.push_back(bseg[b][id[l]][id[r]]);
    }
    assert(((int)(parts).size()) >= 1);
    qans[i] = parts[0];
    for (int j = (1); j < (((int)(parts).size())); ++j)
      qans[i] = merge(qans[i], parts[j]);
  }
}
void run() {
  scanf("%d%d", &n, &nq);
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]), --a[i];
  for (int i = (0); i < (nq); ++i)
    scanf("%d%d", &ql[i], &qr[i]), --ql[i], --qr[i];
  solve();
  printf("%d\n", cnt);
  for (int i = (n); i < (cnt); ++i)
    printf("%d %d\n", how[i].first + 1, how[i].second + 1);
  for (int i = (0); i < (nq); ++i) {
    if (i != 0) printf(" ");
    printf("%d", qans[i] + 1);
  }
  puts("");
}
int main() {
  run();
  return 0;
}