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;
inline void read(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void read(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void readc(char &x) {
  char c;
  while (((c = getchar()) == ' ') || c == '\n')
    ;
  x = c;
}
const int blk = 256;
int n, m, i, j, cnt, lc[2222222], rc[2222222], ans[65545];
struct ii {
  int x, id;
} a[4105];
int merge(int x, int y) {
  if (!x || !y) return x + y;
  cnt++;
  lc[cnt] = x;
  rc[cnt] = y;
  return cnt;
}
struct block {
  vector<vector<int> > id;
  vector<int> pos;
  void init(int x) {
    id.resize(x);
    pos.resize(x);
    int i;
    for (((i)) = (0); ((i)) <= (((int)(x)) - 1); ((i))++) id[i].resize(x - i);
  }
  void build(int x) {
    if (!x)
      init(0);
    else {
      init(1);
      pos[0] = a[x].id;
      id[0][0] = a[x].id;
    }
  }
  int query(int l, int r) {
    l = lower_bound((pos).begin(), (pos).end(), l) - pos.begin();
    r = upper_bound((pos).begin(), (pos).end(), r) - pos.begin() - 1;
    if (l > r) return 0;
    return id[l][r - l];
  }
  void update(block a, block b) {
    init(a.pos.size() + b.pos.size());
    merge(a.pos.begin(), a.pos.end(), b.pos.begin(), b.pos.end(), pos.begin());
    int i, j;
    for (((i)) = (0); ((i)) <= (((int)(pos.size())) - 1); ((i))++)
      for ((j) = (i); (j) <= (pos.size() - 1); (j)++) {
        id[i][j - i] = merge(a.query(pos[i], pos[j]), b.query(pos[i], pos[j]));
      }
  }
} t[17];
bool cmp(ii x, ii y) { return x.x < y.x; }
block solve(int l, int r) {
  if (l == r) {
    block tmp;
    tmp.build(l);
    return tmp;
  }
  int mid = (l + r) / 2;
  block tmp;
  tmp.update(solve(l, mid), solve(mid + 1, r));
  return tmp;
}
int main() {
  read(n);
  read(m);
  cnt = n;
  for (((i)) = (1); ((i)) <= ((n)); ((i))++) {
    read(a[i].x);
    a[i].id = i;
  }
  sort(a + 1, a + n + 1, cmp);
  for (i = 1; i <= n; i += blk) {
    t[i / blk] = solve(i, min(n, i + blk - 1));
  }
  for (((i)) = (1); ((i)) <= ((m)); ((i))++) {
    int l, r;
    read(l);
    read(r);
    for (j = 1; j <= n; j += blk) {
      ans[i] = merge(ans[i], t[j / blk].query(l, r));
    }
  }
  printf("%d\n", cnt);
  for ((i) = (n + 1); (i) <= (cnt); (i)++) printf("%d %d\n", lc[i], rc[i]);
  for (((i)) = (1); ((i)) <= ((m)); ((i))++) printf("%d ", ans[i]);
  return 0;
}