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, S, m;
const int N = 1 << 12, Q = 1 << 16;
int C[1 << 8][N];
int a[N], p[N];
vector<int> B[1 << 8];
int block_index_in_Sid[1 << 8];
vector<vector<vector<int> > > Sid;
int branch_id;
pair<int, int> ans[2200000];
void make_set_in_single_block(const vector<int> &b, int Min, int Max,
                              int nowid) {
  int n = b.size();
  Sid.push_back(vector<vector<int> >(n, vector<int>(n)));
  int mid = (Min + Max) / 2;
  if (Min == Max) {
    Sid[nowid][0][0] = p[b[0]];
    return;
  }
  vector<int> son[2];
  for (int i = 0; i < n; i++) son[b[i] > mid].push_back(b[i]);
  int lson_id = ++branch_id;
  make_set_in_single_block(son[0], Min, mid, lson_id);
  int rson_id = ++branch_id;
  make_set_in_single_block(son[1], mid + 1, Max, rson_id);
  const vector<vector<int> > &lmat = Sid[lson_id];
  const vector<vector<int> > &rmat = Sid[rson_id];
  int icnt[2] = {0, 0};
  int jcnt[2] = {0, 0};
  for (int i = 0; i < n; i++) {
    icnt[b[i] > mid]++;
    jcnt[0] = icnt[0];
    jcnt[1] = icnt[1];
    if (b[i] <= mid) {
      Sid[nowid][i][i] = lmat[icnt[0] - 1][icnt[0] - 1];
    } else {
      Sid[nowid][i][i] = rmat[icnt[1] - 1][icnt[1] - 1];
    }
    for (int j = i + 1; j < n; j++) {
      jcnt[b[j] > mid]++;
      if (jcnt[1] - icnt[1] + (b[i] > mid) == j - i + 1) {
        Sid[nowid][i][j] = rmat[icnt[1] - 1][jcnt[1] - 1];
      } else if (jcnt[0] - icnt[0] + (b[i] <= mid) == j - i + 1) {
        Sid[nowid][i][j] = lmat[icnt[0] - 1][jcnt[0] - 1];
      } else {
        int lL = icnt[0] - (b[i] <= mid), rL = icnt[1] - (b[i] > mid);
        ans[m] = {lmat[lL][jcnt[0] - 1], rmat[rL][jcnt[1] - 1]};
        Sid[nowid][i][j] = m;
        m++;
      }
    }
  }
}
void make_set_in_every_block() {}
void test1() {
  vector<int> a = {1, 0, 2, 3};
  a = {8, 7, 1, 0, 4, 5, 2, 6, 3};
  S = m = n = a.size();
  for (int i = 0; i < n; i++) {
    p[a[i]] = i;
    int block_index = a[i] / S;
    if (i) {
      for (int j = 0; j < S; j++) {
        C[j][i] = C[j][i - 1];
      }
    }
    C[block_index][i]++;
    B[block_index].push_back(a[i]);
  }
  make_set_in_single_block(a, 0, n - 1, branch_id);
  cout << m << endl;
  for (int i = 0; i < m; i++)
    cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << endl;
}
void query() {
  vector<int> k(q);
  for (int t = 0; t < q; t++) {
    int l, r;
    scanf("%d%d", &l, &r);
    --l, --r;
    int last = -1, s = 0;
    for (; s < S; s++) {
      int bR = C[s][r];
      int bL = l ? C[s][l - 1] : 0;
      if (bR - bL > 0) {
        last = Sid[block_index_in_Sid[s]][bL][bR - 1];
      }
      if (last >= 0) break;
    }
    bool update = 0;
    for (s = s + 1; s < S; s++) {
      if (B[s].size() == 0) break;
      int bR = C[s][r];
      int bL = l ? C[s][l - 1] : 0;
      if (bR - bL > 0) {
        update = 1;
        int now = Sid[block_index_in_Sid[s]][bL][bR - 1];
        ans[m] = {last, now};
        last = m;
        m++;
      }
    }
    k[t] = update ? m - 1 : last;
  }
  printf("%d\n", m);
  for (int i = n; i < m; i++)
    printf("%d %d\n", ans[i].first + 1, ans[i].second + 1);
  for (auto x : k) printf("%d ", x + 1);
  puts("");
}
int main() {
  scanf("%d%d", &n, &q);
  S = sqrt(q);
  while (S * S < n) S++;
  m = n;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]), --a[i];
    p[a[i]] = i;
    int block_index = a[i] / S;
    if (i) {
      for (int j = 0; j < S; j++) {
        C[j][i] = C[j][i - 1];
      }
    }
    C[block_index][i]++;
    B[block_index].push_back(a[i]);
  }
  branch_id = -1;
  for (int i = 0; i < S; i++) {
    if (B[i].size() == 0) break;
    branch_id++;
    block_index_in_Sid[i] = branch_id;
    make_set_in_single_block(B[i], i * S, i * S + B[i].size() - 1, branch_id);
  }
  query();
  return 0;
}